Question

A 50.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300 and the coefficient of kinetic friction is 0.200. What is the friction force on the box if (a) a horizontal 140-N push is applied to it?

Answer 1

The friction force on a 50.0-kg box with a 140-N horizontal push applied is 140 N. The push does not overcome the maximum static friction force; hence, the box remains stationary.

Step-by-step explanation:

To determine the friction force on a 50.0-kg box when a horizontal 140-N push is applied, we first need to calculate the maximum static friction force that can act on the box. The static friction force can be calculated using the formula fs = μs × N, where μs is the coefficient of static friction and N is the normal force. The normal force is equal to the weight of the box, which is the mass times the acceleration due to gravity (N = mg).

In this case, N = (50.0 kg)(9.81 m/s2) = 490.5 N. Therefore, the maximum static friction force is fs(max) = (0.300)(490.5 N) = 147.15 N. Since the applied force of 140 N is less than the maximum static friction force, the box does not move, and the force of friction is equal to the applied force. The friction force on the box is 140 N.

Answer 2

98N and 147N

Step-by-step explanation:

We have the following information:

`m=50kg\\\mu_s =0.4\\\mu_k = 0.2\\F=140N`

We can find the static fricton force as follow,

`F=\mu_s * N`

Where N is the normal force (mg)

`F=0.3*50*9.8\\F=147N`

Static friction force at 147N is greater than the force applied hence body does not move.

`F=\mu_k N = 0.2*50*9.8= 98N`