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Found in: Page 143

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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# Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 75.0 N, the second a force of 90.0 N, friction is 12.0 N, and the mass of the third child plus wagon is 23.0 kg. (a) What is the system of interest if the acceleration of the child in the wagon is to be calculated? (b) Draw a free-body diagram, including all forces acting on the system. (c) Calculate the acceleration. (d) What would the acceleration be if friction were 15.0 N?

(a) The system here is the child inside the wagon along with the wagon.

(b) The free-body diagram is drawn as shown below.

(c) The value of acceleration is 0.13 m/s2.

(d) The value of acceleration will be zero.

See the step by step solution

## Step 1: (a) Determine the system of interest

For calculating the acceleration of the child inside the wagon, the system would be the child inside the wagon along with the wagon.

## Step 2: Given data

• Force exerted by the first child of 75.0 N.
• the second child exerts a force of 90.0 N.
• Friction is 12.0 N.
• the mass of the third child plus wagon is 23.0 kg.

## Step 3: (b) Draw the free body diagram of the system considering the forces acting on it

Draw the free-body diagram of the system along with the forces acting on it as shown below,

Here, W is the weight of the wagon plus the child, F1 is the force exerted by the first child, F2 is the force applied by the second child, f is the friction force, N is the normal reaction, m is the mass of wagon plus the third child.

## Step 4: (c) Calculate the acceleration

Apply Newton’s second law of motion.

${F}_{net}=ma\phantom{\rule{0ex}{0ex}}{F}_{2}-{F}_{1}-f=ma................\left(i\right)$

Here, Fnet is the net force, and a is the acceleration.

Substitute 75N for F1, 90N for F2, 12N for f, and 23kg for m in equation (i), and we get,

$90 N-75N-12N=23kg×a\phantom{\rule{0ex}{0ex}}3N=23Kg×a\phantom{\rule{0ex}{0ex}}a=\frac{3kg.m/{s}^{2}}{23kg}\phantom{\rule{0ex}{0ex}}a=0.13m/{s}^{}$

Hence, the value of the acceleration is 0.13 m/s2.

## Step 5: (d) Calculate the acceleration when the force of friction is 15N

Substitute 75N for F1, 90N for F2, 15N for f, and 23kg for m in equation (i), and we get,

$90N-75N-15N=23\mathrm{kg}×a\phantom{\rule{0ex}{0ex}}0=23\mathrm{kg}×a\phantom{\rule{0ex}{0ex}}a=0$

Hence, the acceleration will be 0.

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