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Q9PE

Expert-verifiedFound in: Page 143

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**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/s^{2}.

(d) The value of acceleration will be zero.

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

- 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.

**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, F _{1} is the force exerted by the first child, F_{2} 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.**

Apply Newton’s second law of motion.

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

Here, F_{net} is the net force, and a is the acceleration.

Substitute 75N for F_{1}, 90N for F_{2}, 12N for f, and 23kg for m in equation (i), and we get,

$90\hspace{0.17em}N-75N-12N=23kg\times a\phantom{\rule{0ex}{0ex}}3N=23Kg\times a\phantom{\rule{0ex}{0ex}}a=\frac{3kg.m/{s}^{2}}{23kg}\phantom{\rule{0ex}{0ex}}a=0.13m/{s}^{}$

Hence, the value of the acceleration is 0.13 m/s^{2}.

Substitute 75N for F_{1}, 90N for F_{2}, 15N for f, and 23kg for m in equation (i), and we get,

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

Hence, the acceleration will be 0.

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