The magnitude and direction of the acceleration of pumpkin and the watermelon.

Question

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Answer 1

Question

Chapter 4, Problem 167P

(a)

To determine

The magnitude and direction of the acceleration of pumpkin and the watermelon.

(a)

Expert Solution

Answer to Problem 167P

The magnitude of acceleration of pumpkin and the watermelon is $0.34 m/s2$ the direction of watermelon slides up the ramp and pumpkin slides down.

Explanation of Solution

The figure 1 shows net force exerted on the watermelon and pumpkin attached by the cord

Physics, Chapter 4, Problem 167P

Since the watermelon and pumpkin are attached by the cord, they must have the same magnitude of acceleration.

Write the equation of the net force exerted on the pumpkin by using Newton’s second law.

$∑Fx=mpampgsinθp−T=mpa$ (I)

Here, $mp$ is the mass of the pumpkin, $g$ is the gravitational acceleration, $T$ is the tension exerted by the force on pumpkin attached by the cord, $θp$ is the angle between the pumpkins from the normal line, and $a$ is the acceleration of the pumpkin.

Write the equation of the net force exerted on the watermelon by using Newton’s second law.

$∑Fx=mwaT−mwgsinθw=mwa$ (II)

Here, $mw$ is the mass of the watermelon, $T$ is the tension exerted by the force on watermelon attached by the cord, $θw$ is the angle between the watermelons from the normal line, and $a$ is the acceleration of the watermelon.

Solve and add the equation (I) and (II) for acceleration.

$mpgsinpθ−T=mpaT−mwgsinθw=mwa_g(mpsinθp−mwsinθw)=(mp+mw)aa=g(mpsinθp−mwsinθw)(mp+mw)$

Conclusion:

Substitute $9.80 m/s2$ for $g$ , $7.00 kg$ for $mp$ , $53.0°$ for $θp$ , $10.0 kg$ for $mw$ , and $30.0°$ for $θw$ in above equation.

$a=(9.80 m/s2)[(7.00 kg)sin(53.0°)−(10.0 kg)sin(30.0°)](7.00 kg+10.0 kg)=5.786 kg⋅m/s217.0 kg=0.34 m/s2$

The acceleration is positive, so the direction of watermelon slides up the ramp and pumpkin slides down.

Therefore, the magnitude of acceleration of pumpkin and the watermelon is $0.34 m/s2$ . Here the watermelon moves up and to the left.

(b)

To determine

The distance of the pumpkin after release from the inclined path.

(b)

Expert Solution

Answer to Problem 167P

The distance of the pumpkin after release from the inclined path is $1.5 cm$ .

Explanation of Solution

Write the equation of motion for the pumpkin travel down the ramp.

$Δx=uΔt+12aΔt2$ (III)

Here, $Δx$ is the total distance of the pumpkin slides from the ramp, $u$ is the initial velocity of the pumpkin, $a$ is acceleration of pumpkin, and $Δt$ is the traveling time of the pumpkin.

Conclusion:

Substitute $0 m/s$ for $u$ , $0.30 s$ for $Δt$ , and $0.34 m/s2$ for $a$ in equation (III).

$Δx=(0 m/s)(0.30 s)+12(0.34 m/s2)(0.30 s)2=0.015 m=0.015 m(100 cm1 m)=1.5 cm$

Therefore, the distance of the pumpkin after release from the inclined path is $1.5 cm$ .

(c)

To determine

The speed of the watermelon.

(c)

Expert Solution

Answer to Problem 167P

The speed of the watermelon is $6.8 cm/s$ .

Explanation of Solution

Write the equation of motion for the watermelon gains its speed down the ramp.

$v=u+aΔt$ (IV)

Here, $v$ is the speed of the watermelon, $u$ is the initial velocity of the watermelon, $a$ is acceleration of watermelon, and $Δt$ is the traveling time of the watermelon.

Conclusion:

Substitute $0 m/s$ for $u$ , $0.20 s$ for $Δt$ , and $0.34 m/s2$ for $a$ in equation (IV).

$v=(0 m/s)+(0.34 m/s2)(0.20 s)=0.068 m/s=0.068 m/s(100 cm/s1 m/s)=6.8 cm/s$

Therefore, the speed of the watermelon is $6.8 cm/s$ .

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Answer 2

Question

Chapter 4, Problem 167P

(a)

To determine

The magnitude and direction of the acceleration of pumpkin and the watermelon.

(a)

Expert Solution

Answer to Problem 167P

The magnitude of acceleration of pumpkin and the watermelon is $0.34 m/s2$ the direction of watermelon slides up the ramp and pumpkin slides down.

Explanation of Solution

The figure 1 shows net force exerted on the watermelon and pumpkin attached by the cord

Physics, Chapter 4, Problem 167P

Since the watermelon and pumpkin are attached by the cord, they must have the same magnitude of acceleration.

Write the equation of the net force exerted on the pumpkin by using Newton’s second law.

$∑Fx=mpampgsinθp−T=mpa$ (I)

Here, $mp$ is the mass of the pumpkin, $g$ is the gravitational acceleration, $T$ is the tension exerted by the force on pumpkin attached by the cord, $θp$ is the angle between the pumpkins from the normal line, and $a$ is the acceleration of the pumpkin.

Write the equation of the net force exerted on the watermelon by using Newton’s second law.

$∑Fx=mwaT−mwgsinθw=mwa$ (II)

Here, $mw$ is the mass of the watermelon, $T$ is the tension exerted by the force on watermelon attached by the cord, $θw$ is the angle between the watermelons from the normal line, and $a$ is the acceleration of the watermelon.

Solve and add the equation (I) and (II) for acceleration.

$mpgsinpθ−T=mpaT−mwgsinθw=mwa_g(mpsinθp−mwsinθw)=(mp+mw)aa=g(mpsinθp−mwsinθw)(mp+mw)$

Conclusion:

Substitute $9.80 m/s2$ for $g$ , $7.00 kg$ for $mp$ , $53.0°$ for $θp$ , $10.0 kg$ for $mw$ , and $30.0°$ for $θw$ in above equation.

$a=(9.80 m/s2)[(7.00 kg)sin(53.0°)−(10.0 kg)sin(30.0°)](7.00 kg+10.0 kg)=5.786 kg⋅m/s217.0 kg=0.34 m/s2$

The acceleration is positive, so the direction of watermelon slides up the ramp and pumpkin slides down.

Therefore, the magnitude of acceleration of pumpkin and the watermelon is $0.34 m/s2$ . Here the watermelon moves up and to the left.

(b)

To determine

The distance of the pumpkin after release from the inclined path.

(b)

Expert Solution

Answer to Problem 167P

The distance of the pumpkin after release from the inclined path is $1.5 cm$ .

Explanation of Solution

Write the equation of motion for the pumpkin travel down the ramp.

$Δx=uΔt+12aΔt2$ (III)

Here, $Δx$ is the total distance of the pumpkin slides from the ramp, $u$ is the initial velocity of the pumpkin, $a$ is acceleration of pumpkin, and $Δt$ is the traveling time of the pumpkin.

Conclusion:

Substitute $0 m/s$ for $u$ , $0.30 s$ for $Δt$ , and $0.34 m/s2$ for $a$ in equation (III).

$Δx=(0 m/s)(0.30 s)+12(0.34 m/s2)(0.30 s)2=0.015 m=0.015 m(100 cm1 m)=1.5 cm$

Therefore, the distance of the pumpkin after release from the inclined path is $1.5 cm$ .

(c)

To determine

The speed of the watermelon.

(c)

Expert Solution

Answer to Problem 167P

The speed of the watermelon is $6.8 cm/s$ .

Explanation of Solution

Write the equation of motion for the watermelon gains its speed down the ramp.

$v=u+aΔt$ (IV)

Here, $v$ is the speed of the watermelon, $u$ is the initial velocity of the watermelon, $a$ is acceleration of watermelon, and $Δt$ is the traveling time of the watermelon.

Conclusion:

Substitute $0 m/s$ for $u$ , $0.20 s$ for $Δt$ , and $0.34 m/s2$ for $a$ in equation (IV).

$v=(0 m/s)+(0.34 m/s2)(0.20 s)=0.068 m/s=0.068 m/s(100 cm/s1 m/s)=6.8 cm/s$

Therefore, the speed of the watermelon is $6.8 cm/s$ .

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Answer 3

Question

Chapter 4, Problem 167P

(a)

To determine

The magnitude and direction of the acceleration of pumpkin and the watermelon.

(a)

Expert Solution

Answer to Problem 167P

The magnitude of acceleration of pumpkin and the watermelon is $0.34 m/s2$ the direction of watermelon slides up the ramp and pumpkin slides down.

Explanation of Solution

The figure 1 shows net force exerted on the watermelon and pumpkin attached by the cord

Physics, Chapter 4, Problem 167P

Since the watermelon and pumpkin are attached by the cord, they must have the same magnitude of acceleration.

Write the equation of the net force exerted on the pumpkin by using Newton’s second law.

$∑Fx=mpampgsinθp−T=mpa$ (I)

Here, $mp$ is the mass of the pumpkin, $g$ is the gravitational acceleration, $T$ is the tension exerted by the force on pumpkin attached by the cord, $θp$ is the angle between the pumpkins from the normal line, and $a$ is the acceleration of the pumpkin.

Write the equation of the net force exerted on the watermelon by using Newton’s second law.

$∑Fx=mwaT−mwgsinθw=mwa$ (II)

Here, $mw$ is the mass of the watermelon, $T$ is the tension exerted by the force on watermelon attached by the cord, $θw$ is the angle between the watermelons from the normal line, and $a$ is the acceleration of the watermelon.

Solve and add the equation (I) and (II) for acceleration.

$mpgsinpθ−T=mpaT−mwgsinθw=mwa_g(mpsinθp−mwsinθw)=(mp+mw)aa=g(mpsinθp−mwsinθw)(mp+mw)$

Conclusion:

Substitute $9.80 m/s2$ for $g$ , $7.00 kg$ for $mp$ , $53.0°$ for $θp$ , $10.0 kg$ for $mw$ , and $30.0°$ for $θw$ in above equation.

$a=(9.80 m/s2)[(7.00 kg)sin(53.0°)−(10.0 kg)sin(30.0°)](7.00 kg+10.0 kg)=5.786 kg⋅m/s217.0 kg=0.34 m/s2$

The acceleration is positive, so the direction of watermelon slides up the ramp and pumpkin slides down.

Therefore, the magnitude of acceleration of pumpkin and the watermelon is $0.34 m/s2$ . Here the watermelon moves up and to the left.

(b)

To determine

The distance of the pumpkin after release from the inclined path.

(b)

Expert Solution

Answer to Problem 167P

The distance of the pumpkin after release from the inclined path is $1.5 cm$ .

Explanation of Solution

Write the equation of motion for the pumpkin travel down the ramp.

$Δx=uΔt+12aΔt2$ (III)

Here, $Δx$ is the total distance of the pumpkin slides from the ramp, $u$ is the initial velocity of the pumpkin, $a$ is acceleration of pumpkin, and $Δt$ is the traveling time of the pumpkin.

Conclusion:

Substitute $0 m/s$ for $u$ , $0.30 s$ for $Δt$ , and $0.34 m/s2$ for $a$ in equation (III).

$Δx=(0 m/s)(0.30 s)+12(0.34 m/s2)(0.30 s)2=0.015 m=0.015 m(100 cm1 m)=1.5 cm$

Therefore, the distance of the pumpkin after release from the inclined path is $1.5 cm$ .

(c)

To determine

The speed of the watermelon.

(c)

Expert Solution

Answer to Problem 167P

The speed of the watermelon is $6.8 cm/s$ .

Explanation of Solution

Write the equation of motion for the watermelon gains its speed down the ramp.

$v=u+aΔt$ (IV)

Here, $v$ is the speed of the watermelon, $u$ is the initial velocity of the watermelon, $a$ is acceleration of watermelon, and $Δt$ is the traveling time of the watermelon.

Conclusion:

Substitute $0 m/s$ for $u$ , $0.20 s$ for $Δt$ , and $0.34 m/s2$ for $a$ in equation (IV).

$v=(0 m/s)+(0.34 m/s2)(0.20 s)=0.068 m/s=0.068 m/s(100 cm/s1 m/s)=6.8 cm/s$

Therefore, the speed of the watermelon is $6.8 cm/s$ .

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Answer 4

Question

Chapter 4, Problem 167P

(a)

To determine

The magnitude and direction of the acceleration of pumpkin and the watermelon.

(a)

Expert Solution

Answer to Problem 167P

The magnitude of acceleration of pumpkin and the watermelon is $0.34 m/s2$ the direction of watermelon slides up the ramp and pumpkin slides down.

Explanation of Solution

The figure 1 shows net force exerted on the watermelon and pumpkin attached by the cord

Physics, Chapter 4, Problem 167P

Since the watermelon and pumpkin are attached by the cord, they must have the same magnitude of acceleration.

Write the equation of the net force exerted on the pumpkin by using Newton’s second law.

$∑Fx=mpampgsinθp−T=mpa$ (I)

Here, $mp$ is the mass of the pumpkin, $g$ is the gravitational acceleration, $T$ is the tension exerted by the force on pumpkin attached by the cord, $θp$ is the angle between the pumpkins from the normal line, and $a$ is the acceleration of the pumpkin.

Write the equation of the net force exerted on the watermelon by using Newton’s second law.

$∑Fx=mwaT−mwgsinθw=mwa$ (II)

Here, $mw$ is the mass of the watermelon, $T$ is the tension exerted by the force on watermelon attached by the cord, $θw$ is the angle between the watermelons from the normal line, and $a$ is the acceleration of the watermelon.

Solve and add the equation (I) and (II) for acceleration.

$mpgsinpθ−T=mpaT−mwgsinθw=mwa_g(mpsinθp−mwsinθw)=(mp+mw)aa=g(mpsinθp−mwsinθw)(mp+mw)$

Conclusion:

Substitute $9.80 m/s2$ for $g$ , $7.00 kg$ for $mp$ , $53.0°$ for $θp$ , $10.0 kg$ for $mw$ , and $30.0°$ for $θw$ in above equation.

$a=(9.80 m/s2)[(7.00 kg)sin(53.0°)−(10.0 kg)sin(30.0°)](7.00 kg+10.0 kg)=5.786 kg⋅m/s217.0 kg=0.34 m/s2$

The acceleration is positive, so the direction of watermelon slides up the ramp and pumpkin slides down.

Therefore, the magnitude of acceleration of pumpkin and the watermelon is $0.34 m/s2$ . Here the watermelon moves up and to the left.

(b)

To determine

The distance of the pumpkin after release from the inclined path.

(b)

Expert Solution

Answer to Problem 167P

The distance of the pumpkin after release from the inclined path is $1.5 cm$ .

Explanation of Solution

Write the equation of motion for the pumpkin travel down the ramp.

$Δx=uΔt+12aΔt2$ (III)

Here, $Δx$ is the total distance of the pumpkin slides from the ramp, $u$ is the initial velocity of the pumpkin, $a$ is acceleration of pumpkin, and $Δt$ is the traveling time of the pumpkin.

Conclusion:

Substitute $0 m/s$ for $u$ , $0.30 s$ for $Δt$ , and $0.34 m/s2$ for $a$ in equation (III).

$Δx=(0 m/s)(0.30 s)+12(0.34 m/s2)(0.30 s)2=0.015 m=0.015 m(100 cm1 m)=1.5 cm$

Therefore, the distance of the pumpkin after release from the inclined path is $1.5 cm$ .

(c)

To determine

The speed of the watermelon.

(c)

Expert Solution

Answer to Problem 167P

The speed of the watermelon is $6.8 cm/s$ .

Explanation of Solution

Write the equation of motion for the watermelon gains its speed down the ramp.

$v=u+aΔt$ (IV)

Here, $v$ is the speed of the watermelon, $u$ is the initial velocity of the watermelon, $a$ is acceleration of watermelon, and $Δt$ is the traveling time of the watermelon.

Conclusion:

Substitute $0 m/s$ for $u$ , $0.20 s$ for $Δt$ , and $0.34 m/s2$ for $a$ in equation (IV).

$v=(0 m/s)+(0.34 m/s2)(0.20 s)=0.068 m/s=0.068 m/s(100 cm/s1 m/s)=6.8 cm/s$

Therefore, the speed of the watermelon is $6.8 cm/s$ .

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Answer 5

Chapter 4, Problem 167P

(a)

To determine

The magnitude and direction of the acceleration of pumpkin and the watermelon.

(a)

Expert Solution

Answer to Problem 167P

The magnitude of acceleration of pumpkin and the watermelon is $0.34 m/s2$ the direction of watermelon slides up the ramp and pumpkin slides down.

Explanation of Solution

The figure 1 shows net force exerted on the watermelon and pumpkin attached by the cord

Physics, Chapter 4, Problem 167P

Since the watermelon and pumpkin are attached by the cord, they must have the same magnitude of acceleration.

Write the equation of the net force exerted on the pumpkin by using Newton’s second law.

$∑Fx=mpampgsinθp−T=mpa$ (I)

Here, $mp$ is the mass of the pumpkin, $g$ is the gravitational acceleration, $T$ is the tension exerted by the force on pumpkin attached by the cord, $θp$ is the angle between the pumpkins from the normal line, and $a$ is the acceleration of the pumpkin.

Write the equation of the net force exerted on the watermelon by using Newton’s second law.

$∑Fx=mwaT−mwgsinθw=mwa$ (II)

Here, $mw$ is the mass of the watermelon, $T$ is the tension exerted by the force on watermelon attached by the cord, $θw$ is the angle between the watermelons from the normal line, and $a$ is the acceleration of the watermelon.

Solve and add the equation (I) and (II) for acceleration.

$mpgsinpθ−T=mpaT−mwgsinθw=mwa_g(mpsinθp−mwsinθw)=(mp+mw)aa=g(mpsinθp−mwsinθw)(mp+mw)$

Conclusion:

Substitute $9.80 m/s2$ for $g$ , $7.00 kg$ for $mp$ , $53.0°$ for $θp$ , $10.0 kg$ for $mw$ , and $30.0°$ for $θw$ in above equation.

$a=(9.80 m/s2)[(7.00 kg)sin(53.0°)−(10.0 kg)sin(30.0°)](7.00 kg+10.0 kg)=5.786 kg⋅m/s217.0 kg=0.34 m/s2$

The acceleration is positive, so the direction of watermelon slides up the ramp and pumpkin slides down.

Therefore, the magnitude of acceleration of pumpkin and the watermelon is $0.34 m/s2$ . Here the watermelon moves up and to the left.

(b)

To determine

The distance of the pumpkin after release from the inclined path.

(b)

Expert Solution

Answer to Problem 167P

The distance of the pumpkin after release from the inclined path is $1.5 cm$ .

Explanation of Solution

Write the equation of motion for the pumpkin travel down the ramp.

$Δx=uΔt+12aΔt2$ (III)

Here, $Δx$ is the total distance of the pumpkin slides from the ramp, $u$ is the initial velocity of the pumpkin, $a$ is acceleration of pumpkin, and $Δt$ is the traveling time of the pumpkin.

Conclusion:

Substitute $0 m/s$ for $u$ , $0.30 s$ for $Δt$ , and $0.34 m/s2$ for $a$ in equation (III).

$Δx=(0 m/s)(0.30 s)+12(0.34 m/s2)(0.30 s)2=0.015 m=0.015 m(100 cm1 m)=1.5 cm$

Therefore, the distance of the pumpkin after release from the inclined path is $1.5 cm$ .

(c)

To determine

The speed of the watermelon.

(c)

Expert Solution

Answer to Problem 167P

The speed of the watermelon is $6.8 cm/s$ .

Explanation of Solution

Write the equation of motion for the watermelon gains its speed down the ramp.

$v=u+aΔt$ (IV)

Here, $v$ is the speed of the watermelon, $u$ is the initial velocity of the watermelon, $a$ is acceleration of watermelon, and $Δt$ is the traveling time of the watermelon.

Conclusion:

Substitute $0 m/s$ for $u$ , $0.20 s$ for $Δt$ , and $0.34 m/s2$ for $a$ in equation (IV).

$v=(0 m/s)+(0.34 m/s2)(0.20 s)=0.068 m/s=0.068 m/s(100 cm/s1 m/s)=6.8 cm/s$

Therefore, the speed of the watermelon is $6.8 cm/s$ .

Answer 6

Chapter 4, Problem 167P

(a)

To determine

The magnitude and direction of the acceleration of pumpkin and the watermelon.

(a)

Expert Solution

Answer to Problem 167P

The magnitude of acceleration of pumpkin and the watermelon is $0.34 m/s2$ the direction of watermelon slides up the ramp and pumpkin slides down.

Explanation of Solution

The figure 1 shows net force exerted on the watermelon and pumpkin attached by the cord

Physics, Chapter 4, Problem 167P

Since the watermelon and pumpkin are attached by the cord, they must have the same magnitude of acceleration.

Write the equation of the net force exerted on the pumpkin by using Newton’s second law.

$∑Fx=mpampgsinθp−T=mpa$ (I)

Here, $mp$ is the mass of the pumpkin, $g$ is the gravitational acceleration, $T$ is the tension exerted by the force on pumpkin attached by the cord, $θp$ is the angle between the pumpkins from the normal line, and $a$ is the acceleration of the pumpkin.

Write the equation of the net force exerted on the watermelon by using Newton’s second law.

$∑Fx=mwaT−mwgsinθw=mwa$ (II)

Here, $mw$ is the mass of the watermelon, $T$ is the tension exerted by the force on watermelon attached by the cord, $θw$ is the angle between the watermelons from the normal line, and $a$ is the acceleration of the watermelon.

Solve and add the equation (I) and (II) for acceleration.

$mpgsinpθ−T=mpaT−mwgsinθw=mwa_g(mpsinθp−mwsinθw)=(mp+mw)aa=g(mpsinθp−mwsinθw)(mp+mw)$

Conclusion:

Substitute $9.80 m/s2$ for $g$ , $7.00 kg$ for $mp$ , $53.0°$ for $θp$ , $10.0 kg$ for $mw$ , and $30.0°$ for $θw$ in above equation.

$a=(9.80 m/s2)[(7.00 kg)sin(53.0°)−(10.0 kg)sin(30.0°)](7.00 kg+10.0 kg)=5.786 kg⋅m/s217.0 kg=0.34 m/s2$

The acceleration is positive, so the direction of watermelon slides up the ramp and pumpkin slides down.

Therefore, the magnitude of acceleration of pumpkin and the watermelon is $0.34 m/s2$ . Here the watermelon moves up and to the left.

Answer 7

Chapter 4, Problem 167P

(a)

To determine

The magnitude and direction of the acceleration of pumpkin and the watermelon.

Answer 8

(a)

Expert Solution

Answer to Problem 167P

The magnitude of acceleration of pumpkin and the watermelon is $0.34 m/s2$ the direction of watermelon slides up the ramp and pumpkin slides down.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The figure 1 shows net force exerted on the watermelon and pumpkin attached by the cord

Physics, Chapter 4, Problem 167P

Since the watermelon and pumpkin are attached by the cord, they must have the same magnitude of acceleration.

Write the equation of the net force exerted on the pumpkin by using Newton’s second law.

$∑Fx=mpampgsinθp−T=mpa$ (I)

Here, $mp$ is the mass of the pumpkin, $g$ is the gravitational acceleration, $T$ is the tension exerted by the force on pumpkin attached by the cord, $θp$ is the angle between the pumpkins from the normal line, and $a$ is the acceleration of the pumpkin.

Write the equation of the net force exerted on the watermelon by using Newton’s second law.

$∑Fx=mwaT−mwgsinθw=mwa$ (II)

Here, $mw$ is the mass of the watermelon, $T$ is the tension exerted by the force on watermelon attached by the cord, $θw$ is the angle between the watermelons from the normal line, and $a$ is the acceleration of the watermelon.

Solve and add the equation (I) and (II) for acceleration.

$mpgsinpθ−T=mpaT−mwgsinθw=mwa_g(mpsinθp−mwsinθw)=(mp+mw)aa=g(mpsinθp−mwsinθw)(mp+mw)$

Conclusion:

Substitute $9.80 m/s2$ for $g$ , $7.00 kg$ for $mp$ , $53.0°$ for $θp$ , $10.0 kg$ for $mw$ , and $30.0°$ for $θw$ in above equation.

$a=(9.80 m/s2)[(7.00 kg)sin(53.0°)−(10.0 kg)sin(30.0°)](7.00 kg+10.0 kg)=5.786 kg⋅m/s217.0 kg=0.34 m/s2$

The acceleration is positive, so the direction of watermelon slides up the ramp and pumpkin slides down.

Therefore, the magnitude of acceleration of pumpkin and the watermelon is $0.34 m/s2$ . Here the watermelon moves up and to the left.

Answer 13

(b)

To determine

The distance of the pumpkin after release from the inclined path.

(b)

Expert Solution

Answer to Problem 167P

The distance of the pumpkin after release from the inclined path is $1.5 cm$ .

Explanation of Solution

Write the equation of motion for the pumpkin travel down the ramp.

$Δx=uΔt+12aΔt2$ (III)

Here, $Δx$ is the total distance of the pumpkin slides from the ramp, $u$ is the initial velocity of the pumpkin, $a$ is acceleration of pumpkin, and $Δt$ is the traveling time of the pumpkin.

Conclusion:

Substitute $0 m/s$ for $u$ , $0.30 s$ for $Δt$ , and $0.34 m/s2$ for $a$ in equation (III).

$Δx=(0 m/s)(0.30 s)+12(0.34 m/s2)(0.30 s)2=0.015 m=0.015 m(100 cm1 m)=1.5 cm$

Therefore, the distance of the pumpkin after release from the inclined path is $1.5 cm$ .

Answer 14

(b)

To determine

The distance of the pumpkin after release from the inclined path.

Answer 15

(b)

Expert Solution

Answer to Problem 167P

The distance of the pumpkin after release from the inclined path is $1.5 cm$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the equation of motion for the pumpkin travel down the ramp.

$Δx=uΔt+12aΔt2$ (III)

Here, $Δx$ is the total distance of the pumpkin slides from the ramp, $u$ is the initial velocity of the pumpkin, $a$ is acceleration of pumpkin, and $Δt$ is the traveling time of the pumpkin.

Conclusion:

Substitute $0 m/s$ for $u$ , $0.30 s$ for $Δt$ , and $0.34 m/s2$ for $a$ in equation (III).

$Δx=(0 m/s)(0.30 s)+12(0.34 m/s2)(0.30 s)2=0.015 m=0.015 m(100 cm1 m)=1.5 cm$

Therefore, the distance of the pumpkin after release from the inclined path is $1.5 cm$ .

Answer 20

(c)

To determine

The speed of the watermelon.

(c)

Expert Solution

Answer to Problem 167P

The speed of the watermelon is $6.8 cm/s$ .

Explanation of Solution

Write the equation of motion for the watermelon gains its speed down the ramp.

$v=u+aΔt$ (IV)

Here, $v$ is the speed of the watermelon, $u$ is the initial velocity of the watermelon, $a$ is acceleration of watermelon, and $Δt$ is the traveling time of the watermelon.

Conclusion:

Substitute $0 m/s$ for $u$ , $0.20 s$ for $Δt$ , and $0.34 m/s2$ for $a$ in equation (IV).

$v=(0 m/s)+(0.34 m/s2)(0.20 s)=0.068 m/s=0.068 m/s(100 cm/s1 m/s)=6.8 cm/s$

Therefore, the speed of the watermelon is $6.8 cm/s$ .

Answer 21

(c)

To determine

The speed of the watermelon.

Answer 22

(c)

Expert Solution

Answer to Problem 167P

The speed of the watermelon is $6.8 cm/s$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Write the equation of motion for the watermelon gains its speed down the ramp.

$v=u+aΔt$ (IV)

Here, $v$ is the speed of the watermelon, $u$ is the initial velocity of the watermelon, $a$ is acceleration of watermelon, and $Δt$ is the traveling time of the watermelon.