Question

Suppose two children push horizontally, but in exactly opposite directions, on a third child in a sled. The first child exerts a force of 79 N, the second a force of 92 N, kinetic friction is 5.5 N, and the mass of the third child plus sled is 24 kg.

1. Using a coordinate system where the second child is pushing in the positive direction, calculate the acceleration in m/s2.
2. What is the system of interest if the accelaration of the child in the wagon is to be calculated?
3. Draw a free body diagram including all bodies acting on the system
4. What would be the acceleration if friction were 150 N?

Answer 1

Please, read the anser below

Step-by-step explanation:

1. In order to calculate the acceleration of the children you use the Newton second law for the summation of the implied forces:

`F_2-F_1-F_f=Ma` (1)

Where is has been used that the motion is in the direction of the applied force by the second child

F2: force of the second child = 92N

F1: force of the first child = 79N

Ff: friction force = 5.5N

M: mass of the third child = 24kg

a: acceleration of the third child = ?

You solve the equation (1) for a, and you replace the values of the other parameters:

`a=(F_2-F_1.F_f)/(M)=(96N-79N-5.5N)/(24kg)=0.48(m)/(s^2)`

The acceleration is 0.48m/s^2

2. The system of interest is the same as before, the acceleration calculated is about the motion of the third child.

3. An image with the diagram forces is attached below.

4. If the friction would be 150N, the acceleration would be zero, because the friction force is higher than the higher force between children, which is 92N.

Then, the acceleration is zero