In each case, determine its velocity when t = 2 s if v = 0 when t = 0.

Question

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Answer 1

Textbook Question

Chapter 13.4, Problem 1PP

In each case, determine its velocity when t = 2 s if v = 0 when t = 0. Chapter 13.4, Problem 1PP, In each case, determine its velocity when t = 2 s if v = 0 when t = 0.

Expert Solution

To determine

(a)

The velocity of the block when $t=2 s$ .

Answer to Problem 1PP

The velocity of the block at the time $t=2 s$ is $20 m/s$ .

Explanation of Solution

Given:

The mass of the block is $m=10 kg$ .

Initially the velocity of the block is $v=0$ at $t=0$ .

The free body diagram of the block as shown in Figure (1a).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 1

Write the formula for Newton’s second law of motion along $x−axis$ .

$∑Fx=max$ (I)

Here, $m$ is the mass of the block, $ax$ is the acceleration of the block along $x−axis$ and $∑Fx$ is the resultant of all the forces acting on the block along $x−axis$ .

Write the formula for velocity of the block at constant acceleration as a function of time.

$v=v0+act$ (II)

Here, $v$ is the final velocity, $v0$ is the initial velocity, $ac$ is the constant acceleration and $t$ is the time.

Conclusion:

Refer Figure (1a).

Resolve the forces along $x−axis$ .

$∑Fx=500 N(45)−300 N=100 N$

Calculate the acceleration of the block.

Substitute $100 N$ for $∑Fx$ , $10 kg$ for $m$ in Equation (I).

$100 N=(10 kg)axax=10010ax=10 m/s2$

Calculate the velocity of the block at the time $t= 2 s$ .

Here, $ax=ac=10 m/s2$ .

Substitute $0$ for $v0$ , $10 m/s2$ for $ac$ and $2 s$ for $t$ in Equation (II).

$v=0+(10 m/s2)(2 s)=0+20=20 m/s$

Thus, the velocity of the block at the time $t=2 s$ is $20 m/s$ .

Expert Solution

To determine

b)

The velocity of the block when $t=2 s$ .

Answer to Problem 1PP

The velocity of the block at the time $t=2 s$ is $4 m/s$ .

Explanation of Solution

Given:

The mass of the block is $m=10 kg$ .

Initially the velocity of the block is $v=0$ at $t=0$ .

The free body diagram of the block as shown in Figure (1b).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 2

Write the formula for Newton’s second law of motion along $x−axis$ .

$∑Fx=max$ (I)

Here, $m$ is the mass of the block, $ax$ is the acceleration of the block along $x−axis$ and $∑Fx$ is the resultant of all the forces acting on the block along $x−axis$ .

Write the formula for acceleration $(a)$ of the block.

$a=dvdtdv=a dt$ (II)

Here, $dvdt$ is the rate of change of velocity with respect to time.

Conclusion:

Refer Figure (1b).

Resolve the forces along $x−axis$ .

$∑Fx=20t= 20t$

Calculate the acceleration of the block.

Substitute $20t$ for $∑Fx$ , $10 kg$ for $m$ in Equation (I).

$20t=(10 kg)axax=20t10ax=2t$

Calculate the velocity of the block at the time $t= 2 s$ .

Here, $ax=a=2t$ .

Substitute $2t$ for $a$ in Equation (II).

$dv=2t dt$

Integrate at the limits of $v=0 to v$ and $t=0 to 2 s$ .

$∫0vdv=∫022t dt[v]0v=[2t22]02(v−0)=(2)2−(0)0v=4 m/s$

Thus, the velocity of the block at the time $t=2 s$ is $4 m/s$ .

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Answer 2

Textbook Question

Chapter 13.4, Problem 1PP

In each case, determine its velocity when t = 2 s if v = 0 when t = 0. Chapter 13.4, Problem 1PP, In each case, determine its velocity when t = 2 s if v = 0 when t = 0.

Expert Solution

To determine

(a)

The velocity of the block when $t=2 s$ .

Answer to Problem 1PP

The velocity of the block at the time $t=2 s$ is $20 m/s$ .

Explanation of Solution

Given:

The mass of the block is $m=10 kg$ .

Initially the velocity of the block is $v=0$ at $t=0$ .

The free body diagram of the block as shown in Figure (1a).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 1

Write the formula for Newton’s second law of motion along $x−axis$ .

$∑Fx=max$ (I)

Here, $m$ is the mass of the block, $ax$ is the acceleration of the block along $x−axis$ and $∑Fx$ is the resultant of all the forces acting on the block along $x−axis$ .

Write the formula for velocity of the block at constant acceleration as a function of time.

$v=v0+act$ (II)

Here, $v$ is the final velocity, $v0$ is the initial velocity, $ac$ is the constant acceleration and $t$ is the time.

Conclusion:

Refer Figure (1a).

Resolve the forces along $x−axis$ .

$∑Fx=500 N(45)−300 N=100 N$

Calculate the acceleration of the block.

Substitute $100 N$ for $∑Fx$ , $10 kg$ for $m$ in Equation (I).

$100 N=(10 kg)axax=10010ax=10 m/s2$

Calculate the velocity of the block at the time $t= 2 s$ .

Here, $ax=ac=10 m/s2$ .

Substitute $0$ for $v0$ , $10 m/s2$ for $ac$ and $2 s$ for $t$ in Equation (II).

$v=0+(10 m/s2)(2 s)=0+20=20 m/s$

Thus, the velocity of the block at the time $t=2 s$ is $20 m/s$ .

Expert Solution

To determine

b)

The velocity of the block when $t=2 s$ .

Answer to Problem 1PP

The velocity of the block at the time $t=2 s$ is $4 m/s$ .

Explanation of Solution

Given:

The mass of the block is $m=10 kg$ .

Initially the velocity of the block is $v=0$ at $t=0$ .

The free body diagram of the block as shown in Figure (1b).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 2

Write the formula for Newton’s second law of motion along $x−axis$ .

$∑Fx=max$ (I)

Here, $m$ is the mass of the block, $ax$ is the acceleration of the block along $x−axis$ and $∑Fx$ is the resultant of all the forces acting on the block along $x−axis$ .

Write the formula for acceleration $(a)$ of the block.

$a=dvdtdv=a dt$ (II)

Here, $dvdt$ is the rate of change of velocity with respect to time.

Conclusion:

Refer Figure (1b).

Resolve the forces along $x−axis$ .

$∑Fx=20t= 20t$

Calculate the acceleration of the block.

Substitute $20t$ for $∑Fx$ , $10 kg$ for $m$ in Equation (I).

$20t=(10 kg)axax=20t10ax=2t$

Calculate the velocity of the block at the time $t= 2 s$ .

Here, $ax=a=2t$ .

Substitute $2t$ for $a$ in Equation (II).

$dv=2t dt$

Integrate at the limits of $v=0 to v$ and $t=0 to 2 s$ .

$∫0vdv=∫022t dt[v]0v=[2t22]02(v−0)=(2)2−(0)0v=4 m/s$

Thus, the velocity of the block at the time $t=2 s$ is $4 m/s$ .

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Answer 3

Textbook Question

Chapter 13.4, Problem 1PP

In each case, determine its velocity when t = 2 s if v = 0 when t = 0. Chapter 13.4, Problem 1PP, In each case, determine its velocity when t = 2 s if v = 0 when t = 0.

Expert Solution

To determine

(a)

The velocity of the block when $t=2 s$ .

Answer to Problem 1PP

The velocity of the block at the time $t=2 s$ is $20 m/s$ .

Explanation of Solution

Given:

The mass of the block is $m=10 kg$ .

Initially the velocity of the block is $v=0$ at $t=0$ .

The free body diagram of the block as shown in Figure (1a).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 1

Write the formula for Newton’s second law of motion along $x−axis$ .

$∑Fx=max$ (I)

Here, $m$ is the mass of the block, $ax$ is the acceleration of the block along $x−axis$ and $∑Fx$ is the resultant of all the forces acting on the block along $x−axis$ .

Write the formula for velocity of the block at constant acceleration as a function of time.

$v=v0+act$ (II)

Here, $v$ is the final velocity, $v0$ is the initial velocity, $ac$ is the constant acceleration and $t$ is the time.

Conclusion:

Refer Figure (1a).

Resolve the forces along $x−axis$ .

$∑Fx=500 N(45)−300 N=100 N$

Calculate the acceleration of the block.

Substitute $100 N$ for $∑Fx$ , $10 kg$ for $m$ in Equation (I).

$100 N=(10 kg)axax=10010ax=10 m/s2$

Calculate the velocity of the block at the time $t= 2 s$ .

Here, $ax=ac=10 m/s2$ .

Substitute $0$ for $v0$ , $10 m/s2$ for $ac$ and $2 s$ for $t$ in Equation (II).

$v=0+(10 m/s2)(2 s)=0+20=20 m/s$

Thus, the velocity of the block at the time $t=2 s$ is $20 m/s$ .

Expert Solution

To determine

b)

The velocity of the block when $t=2 s$ .

Answer to Problem 1PP

The velocity of the block at the time $t=2 s$ is $4 m/s$ .

Explanation of Solution

Given:

The mass of the block is $m=10 kg$ .

Initially the velocity of the block is $v=0$ at $t=0$ .

The free body diagram of the block as shown in Figure (1b).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 2

Write the formula for Newton’s second law of motion along $x−axis$ .

$∑Fx=max$ (I)

Here, $m$ is the mass of the block, $ax$ is the acceleration of the block along $x−axis$ and $∑Fx$ is the resultant of all the forces acting on the block along $x−axis$ .

Write the formula for acceleration $(a)$ of the block.

$a=dvdtdv=a dt$ (II)

Here, $dvdt$ is the rate of change of velocity with respect to time.

Conclusion:

Refer Figure (1b).

Resolve the forces along $x−axis$ .

$∑Fx=20t= 20t$

Calculate the acceleration of the block.

Substitute $20t$ for $∑Fx$ , $10 kg$ for $m$ in Equation (I).

$20t=(10 kg)axax=20t10ax=2t$

Calculate the velocity of the block at the time $t= 2 s$ .

Here, $ax=a=2t$ .

Substitute $2t$ for $a$ in Equation (II).

$dv=2t dt$

Integrate at the limits of $v=0 to v$ and $t=0 to 2 s$ .

$∫0vdv=∫022t dt[v]0v=[2t22]02(v−0)=(2)2−(0)0v=4 m/s$

Thus, the velocity of the block at the time $t=2 s$ is $4 m/s$ .

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Answer 4

Textbook Question

Chapter 13.4, Problem 1PP

In each case, determine its velocity when t = 2 s if v = 0 when t = 0. Chapter 13.4, Problem 1PP, In each case, determine its velocity when t = 2 s if v = 0 when t = 0.

Expert Solution

To determine

(a)

The velocity of the block when $t=2 s$ .

Answer to Problem 1PP

The velocity of the block at the time $t=2 s$ is $20 m/s$ .

Explanation of Solution

Given:

The mass of the block is $m=10 kg$ .

Initially the velocity of the block is $v=0$ at $t=0$ .

The free body diagram of the block as shown in Figure (1a).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 1

Write the formula for Newton’s second law of motion along $x−axis$ .

$∑Fx=max$ (I)

Here, $m$ is the mass of the block, $ax$ is the acceleration of the block along $x−axis$ and $∑Fx$ is the resultant of all the forces acting on the block along $x−axis$ .

Write the formula for velocity of the block at constant acceleration as a function of time.

$v=v0+act$ (II)

Here, $v$ is the final velocity, $v0$ is the initial velocity, $ac$ is the constant acceleration and $t$ is the time.

Conclusion:

Refer Figure (1a).

Resolve the forces along $x−axis$ .

$∑Fx=500 N(45)−300 N=100 N$

Calculate the acceleration of the block.

Substitute $100 N$ for $∑Fx$ , $10 kg$ for $m$ in Equation (I).

$100 N=(10 kg)axax=10010ax=10 m/s2$

Calculate the velocity of the block at the time $t= 2 s$ .

Here, $ax=ac=10 m/s2$ .

Substitute $0$ for $v0$ , $10 m/s2$ for $ac$ and $2 s$ for $t$ in Equation (II).

$v=0+(10 m/s2)(2 s)=0+20=20 m/s$

Thus, the velocity of the block at the time $t=2 s$ is $20 m/s$ .

Expert Solution

To determine

b)

The velocity of the block when $t=2 s$ .

Answer to Problem 1PP

The velocity of the block at the time $t=2 s$ is $4 m/s$ .

Explanation of Solution

Given:

The mass of the block is $m=10 kg$ .

Initially the velocity of the block is $v=0$ at $t=0$ .

The free body diagram of the block as shown in Figure (1b).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 2

Write the formula for Newton’s second law of motion along $x−axis$ .

$∑Fx=max$ (I)

Here, $m$ is the mass of the block, $ax$ is the acceleration of the block along $x−axis$ and $∑Fx$ is the resultant of all the forces acting on the block along $x−axis$ .

Write the formula for acceleration $(a)$ of the block.

$a=dvdtdv=a dt$ (II)

Here, $dvdt$ is the rate of change of velocity with respect to time.

Conclusion:

Refer Figure (1b).

Resolve the forces along $x−axis$ .

$∑Fx=20t= 20t$

Calculate the acceleration of the block.

Substitute $20t$ for $∑Fx$ , $10 kg$ for $m$ in Equation (I).

$20t=(10 kg)axax=20t10ax=2t$

Calculate the velocity of the block at the time $t= 2 s$ .

Here, $ax=a=2t$ .

Substitute $2t$ for $a$ in Equation (II).

$dv=2t dt$

Integrate at the limits of $v=0 to v$ and $t=0 to 2 s$ .

$∫0vdv=∫022t dt[v]0v=[2t22]02(v−0)=(2)2−(0)0v=4 m/s$

Thus, the velocity of the block at the time $t=2 s$ is $4 m/s$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

To determine

(a)

The velocity of the block when $t=2 s$ .

Answer to Problem 1PP

The velocity of the block at the time $t=2 s$ is $20 m/s$ .

Explanation of Solution

Given:

The mass of the block is $m=10 kg$ .

Initially the velocity of the block is $v=0$ at $t=0$ .

The free body diagram of the block as shown in Figure (1a).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 1

Write the formula for Newton’s second law of motion along $x−axis$ .

$∑Fx=max$ (I)

Here, $m$ is the mass of the block, $ax$ is the acceleration of the block along $x−axis$ and $∑Fx$ is the resultant of all the forces acting on the block along $x−axis$ .

Write the formula for velocity of the block at constant acceleration as a function of time.

$v=v0+act$ (II)

Here, $v$ is the final velocity, $v0$ is the initial velocity, $ac$ is the constant acceleration and $t$ is the time.

Conclusion:

Refer Figure (1a).

Resolve the forces along $x−axis$ .

$∑Fx=500 N(45)−300 N=100 N$

Calculate the acceleration of the block.

Substitute $100 N$ for $∑Fx$ , $10 kg$ for $m$ in Equation (I).

$100 N=(10 kg)axax=10010ax=10 m/s2$

Calculate the velocity of the block at the time $t= 2 s$ .

Here, $ax=ac=10 m/s2$ .

Substitute $0$ for $v0$ , $10 m/s2$ for $ac$ and $2 s$ for $t$ in Equation (II).

$v=0+(10 m/s2)(2 s)=0+20=20 m/s$

Thus, the velocity of the block at the time $t=2 s$ is $20 m/s$ .

Expert Solution

To determine

b)

The velocity of the block when $t=2 s$ .

Answer to Problem 1PP

The velocity of the block at the time $t=2 s$ is $4 m/s$ .

Explanation of Solution

Given:

The mass of the block is $m=10 kg$ .

Initially the velocity of the block is $v=0$ at $t=0$ .

The free body diagram of the block as shown in Figure (1b).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 2

Write the formula for Newton’s second law of motion along $x−axis$ .

$∑Fx=max$ (I)

Here, $m$ is the mass of the block, $ax$ is the acceleration of the block along $x−axis$ and $∑Fx$ is the resultant of all the forces acting on the block along $x−axis$ .

Write the formula for acceleration $(a)$ of the block.

$a=dvdtdv=a dt$ (II)

Here, $dvdt$ is the rate of change of velocity with respect to time.

Conclusion:

Refer Figure (1b).

Resolve the forces along $x−axis$ .

$∑Fx=20t= 20t$

Calculate the acceleration of the block.

Substitute $20t$ for $∑Fx$ , $10 kg$ for $m$ in Equation (I).

$20t=(10 kg)axax=20t10ax=2t$

Calculate the velocity of the block at the time $t= 2 s$ .

Here, $ax=a=2t$ .

Substitute $2t$ for $a$ in Equation (II).

$dv=2t dt$

Integrate at the limits of $v=0 to v$ and $t=0 to 2 s$ .

$∫0vdv=∫022t dt[v]0v=[2t22]02(v−0)=(2)2−(0)0v=4 m/s$

Thus, the velocity of the block at the time $t=2 s$ is $4 m/s$ .

Answer 8

Expert Solution

To determine

(a)

The velocity of the block when $t=2 s$ .

Answer to Problem 1PP

The velocity of the block at the time $t=2 s$ is $20 m/s$ .

Explanation of Solution

Given:

The mass of the block is $m=10 kg$ .

Initially the velocity of the block is $v=0$ at $t=0$ .

The free body diagram of the block as shown in Figure (1a).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 1

Write the formula for Newton’s second law of motion along $x−axis$ .

$∑Fx=max$ (I)

Here, $m$ is the mass of the block, $ax$ is the acceleration of the block along $x−axis$ and $∑Fx$ is the resultant of all the forces acting on the block along $x−axis$ .

Write the formula for velocity of the block at constant acceleration as a function of time.

$v=v0+act$ (II)

Here, $v$ is the final velocity, $v0$ is the initial velocity, $ac$ is the constant acceleration and $t$ is the time.

Conclusion:

Refer Figure (1a).

Resolve the forces along $x−axis$ .

$∑Fx=500 N(45)−300 N=100 N$

Calculate the acceleration of the block.

Substitute $100 N$ for $∑Fx$ , $10 kg$ for $m$ in Equation (I).

$100 N=(10 kg)axax=10010ax=10 m/s2$

Calculate the velocity of the block at the time $t= 2 s$ .

Here, $ax=ac=10 m/s2$ .

Substitute $0$ for $v0$ , $10 m/s2$ for $ac$ and $2 s$ for $t$ in Equation (II).

$v=0+(10 m/s2)(2 s)=0+20=20 m/s$

Thus, the velocity of the block at the time $t=2 s$ is $20 m/s$ .

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

(a)

The velocity of the block when $t=2 s$ .

Answer 13

Answer to Problem 1PP

The velocity of the block at the time $t=2 s$ is $20 m/s$ .

Answer 14

Explanation of Solution

Given:

The mass of the block is $m=10 kg$ .

Initially the velocity of the block is $v=0$ at $t=0$ .

The free body diagram of the block as shown in Figure (1a).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 1

Write the formula for Newton’s second law of motion along $x−axis$ .

$∑Fx=max$ (I)

Here, $m$ is the mass of the block, $ax$ is the acceleration of the block along $x−axis$ and $∑Fx$ is the resultant of all the forces acting on the block along $x−axis$ .

Write the formula for velocity of the block at constant acceleration as a function of time.

$v=v0+act$ (II)

Here, $v$ is the final velocity, $v0$ is the initial velocity, $ac$ is the constant acceleration and $t$ is the time.

Conclusion:

Refer Figure (1a).

Resolve the forces along $x−axis$ .

$∑Fx=500 N(45)−300 N=100 N$

Calculate the acceleration of the block.

Substitute $100 N$ for $∑Fx$ , $10 kg$ for $m$ in Equation (I).

$100 N=(10 kg)axax=10010ax=10 m/s2$

Calculate the velocity of the block at the time $t= 2 s$ .

Here, $ax=ac=10 m/s2$ .

Substitute $0$ for $v0$ , $10 m/s2$ for $ac$ and $2 s$ for $t$ in Equation (II).

$v=0+(10 m/s2)(2 s)=0+20=20 m/s$

Thus, the velocity of the block at the time $t=2 s$ is $20 m/s$ .

Answer 15

Expert Solution

To determine

b)

The velocity of the block when $t=2 s$ .

Answer to Problem 1PP

The velocity of the block at the time $t=2 s$ is $4 m/s$ .

Explanation of Solution

Given:

The mass of the block is $m=10 kg$ .

Initially the velocity of the block is $v=0$ at $t=0$ .

The free body diagram of the block as shown in Figure (1b).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 2

Write the formula for Newton’s second law of motion along $x−axis$ .

$∑Fx=max$ (I)

Here, $m$ is the mass of the block, $ax$ is the acceleration of the block along $x−axis$ and $∑Fx$ is the resultant of all the forces acting on the block along $x−axis$ .

Write the formula for acceleration $(a)$ of the block.

$a=dvdtdv=a dt$ (II)

Here, $dvdt$ is the rate of change of velocity with respect to time.

Conclusion:

Refer Figure (1b).

Resolve the forces along $x−axis$ .

$∑Fx=20t= 20t$

Calculate the acceleration of the block.

Substitute $20t$ for $∑Fx$ , $10 kg$ for $m$ in Equation (I).

$20t=(10 kg)axax=20t10ax=2t$

Calculate the velocity of the block at the time $t= 2 s$ .

Here, $ax=a=2t$ .

Substitute $2t$ for $a$ in Equation (II).

$dv=2t dt$

Integrate at the limits of $v=0 to v$ and $t=0 to 2 s$ .

$∫0vdv=∫022t dt[v]0v=[2t22]02(v−0)=(2)2−(0)0v=4 m/s$

Thus, the velocity of the block at the time $t=2 s$ is $4 m/s$ .

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

b)

The velocity of the block when $t=2 s$ .

Answer 20

Answer to Problem 1PP

The velocity of the block at the time $t=2 s$ is $4 m/s$ .

Answer 21

Explanation of Solution

Given:

The mass of the block is $m=10 kg$ .

Initially the velocity of the block is $v=0$ at $t=0$ .

The free body diagram of the block as shown in Figure (1b).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 2

Write the formula for Newton’s second law of motion along $x−axis$ .

$∑Fx=max$ (I)

Here, $m$ is the mass of the block, $ax$ is the acceleration of the block along $x−axis$ and $∑Fx$ is the resultant of all the forces acting on the block along $x−axis$ .