Physics, 29.06.2019 16:30 emmacox069

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?

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Answer from: goosenar

a) Child plus wagon

b) Refer the attached diagram.

c) $\rm a = 0.13 \; m/sec^2$

d) a = 0

Given :

The first child exerts a force of 75 N, the second a force of 90 N.

Friction = 12N

Mass of the third child plus wagon is 23 kg.

Solution :

a) According to Newton's second law,

F = ma

Here we are interested in calculating the acceleration (a) of the child. However, the child moves together with the wagon - this means that we can consider the child plus wagon as a single system, moving with an acceleration of a, under a net force of F, which is the result of all the forces applied to the child plus wagon.

b) Refer the attached Diagram.

c) F = ma

90 - 75 - 12 = 23a

$\rm a = \dfrac{3}{23}= 0.13\;m/sec^2$

d) F = ma

90 - 75 - 15 = 23a

a = 0

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Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wa

Answer from: livesk8923

(a) The child + the wagon

Explanation:

Newton's second law states that the net force applied on a system is equal to the mass of the system times its acceleration:

$\sum F = ma$ (1)

Here we are interested in calculating the acceleration of the child, a. However, the child moves together with the wagon - this means that we can consider the child+wagon as a single system, moving with an acceleration of a, under a net force of F, which is the result of all the forces applied to the child+wagon.

(b) See attachment

There are 5 forces in total acting on the child+wagon system:

- The force exerted by the first child, 75.0 N, here to the left

- The force exerted by the second child, 90.0 N, here to the right

- The frictional force of 12.0 N, to the left (the direction of the frictional force is opposite to the direction of motion; since the larger force is the one exerted to the right, the system would move to the right, so the frictional force acts to the left)

- The weight of the child+wagon system, downward, which is equal to

W = mg

where m=23.0 kg is the total mass and g=9.8 m/s^2 is the acceleration due to gravity

- The normal reaction of the surface, upward, equal in magnitude to the weight

We are only interested in the forces acting along the horizontal direction, since the two forces in the vertical direction are in equilibrium.

The resultant force along the horizontal direction is

$\sum F = 90.0 N - 75.0 N - 12.0 N =3.0 N$

By using Newton's second law (eq. 1), we can find the acceleration:

$a=\frac{\sum F}{m}$

where

m = 23.0 kg is the mass of the child+wagon

Substituting,

$a=\frac{3.0 N}{23.0 kg}=0.13 m/s^2$

(d) 0

In this case, the magnitude of the friction is 15.0 N. This means that the net force acting on the system is

$\sum F = 90.0 N -75.0 N -15.0 N =0$

So, the forces along the horizontal direction are balanced as well. According to eq.(1), this also means that the acceleration along the horizontal axis will also be zero:

$a=\frac{0}{23.0 kg}=0$

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