Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 4, Problem 167P

(a)

To determine

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

(a)

Expert Solution
Check Mark

Answer to Problem 167P

The magnitude of acceleration of pumpkin and the watermelon is 0.34m/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=mpampgsinθpT=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=mwaTmwgsinθw=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.

mpgsinpθT=mpaTmwgsinθw=mwa_g(mpsinθpmwsinθw)=(mp+mw)aa=g(mpsinθpmwsinθw)(mp+mw)

Conclusion:

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

a=(9.80m/s2)[(7.00kg)sin(53.0°)(10.0kg)sin(30.0°)](7.00kg+10.0kg)=5.786kgm/s217.0kg=0.34m/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 0.34m/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
Check Mark

Answer to Problem 167P

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

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

Δx=(0m/s)(0.30s)+12(0.34m/s2)(0.30s)2=0.015m=0.015m(100cm1m)=1.5cm

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

(c)

To determine

The speed of the watermelon.

(c)

Expert Solution
Check Mark

Answer to Problem 167P

The speed of the watermelon is 6.8cm/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 0m/s for u, 0.20s for Δt, and 0.34m/s2 for a in equation (IV).

v=(0m/s)+(0.34m/s2)(0.20s)=0.068m/s=0.068m/s(100cm/s1m/s)=6.8cm/s

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

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Chapter 4 Solutions

Physics

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Pulleys and inclined planes are examples of...Ch. 4 - Prob. 25CQCh. 4 - Prob. 26CQCh. 4 - Prob. 27CQCh. 4 - Prob. 28CQCh. 4 - Prob. 29CQCh. 4 - Prob. 30CQCh. 4 - Prob. 31CQCh. 4 - Prob. 32CQCh. 4 - Prob. 1MCQCh. 4 - Prob. 2MCQCh. 4 - Prob. 3MCQCh. 4 - Prob. 4MCQCh. 4 - Prob. 5MCQCh. 4 - Prob. 6MCQCh. 4 - Prob. 7MCQCh. 4 - Prob. 8MCQCh. 4 - Prob. 9MCQCh. 4 - Prob. 10MCQCh. 4 - Prob. 11MCQCh. 4 - Prob. 12MCQCh. 4 - Prob. 13MCQCh. 4 - Prob. 14MCQCh. 4 - Prob. 15MCQCh. 4 - Prob. 16MCQCh. 4 - Prob. 17MCQCh. 4 - Prob. 18MCQCh. 4 - Prob. 19MCQCh. 4 - Prob. 20MCQCh. 4 - Prob. 21MCQCh. 4 - Prob. 22MCQCh. 4 - Prob. 23MCQCh. 4 - Prob. 24MCQCh. 4 - Prob. 25MCQCh. 4 - Prob. 26MCQCh. 4 - Prob. 1PCh. 4 - Prob. 2PCh. 4 - Prob. 3PCh. 4 - Prob. 4PCh. 4 - Prob. 5PCh. 4 - Prob. 6PCh. 4 - Prob. 7PCh. 4 - Prob. 8PCh. 4 - Prob. 9PCh. 4 - Prob. 10PCh. 4 - Prob. 11PCh. 4 - Prob. 12PCh. 4 - Prob. 13PCh. 4 - Prob. 14PCh. 4 - Prob. 15PCh. 4 - 16. A truck driving on a level highway is acted on...Ch. 4 - 17. 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