Forces and velocities of a sliding box

[tiffanyg]

1. a box of mass 12 kg slides at a speed of 10 m/s acros a smooth level floor, where it enters a rough portion 3.0 m in length. In the rough portion, the box experiences a horizontal friction force of 72N.

a) how much work is done by the frictional force?
b) what is the velocity of the box when it leaves the rough surface?
c) what length of rough surface brings the box coompletely to rest?


I don't know how to do it!

 

[SeanGillespie]

a) What is the equation for work done? What you need to calculate should be quite obvious.

b) You are given the force of friction, what is the equation for Force? You should be able to calculate what you need to calculate the reduced velocity (you will need a SUVAT equation here also).

See if that helps at all... check back with your answers...

 

[kerimek]

I can help you understand the forces and velocities involved in this scenario. Let's break down the problem into smaller parts and use some basic physics principles to find the answers.

First, we need to understand that an object in motion will continue to move at a constant velocity unless acted upon by an external force. In this case, the box has a mass of 12 kg and a velocity of 10 m/s, so we can use the equation F=ma to calculate the net force acting on the box. The net force is the sum of all the forces acting on the box.

a) We are given that the box experiences a friction force of 72N in the rough portion. We can use this information to calculate the work done by the frictional force using the equation W=Fd, where W is work, F is the force, and d is the distance. In this case, the work done by the frictional force is 72N x 3.0m = 216J.

b) To find the velocity of the box when it leaves the rough surface, we can use the principle of conservation of energy. This principle states that the total energy of a system remains constant, and energy can only be transferred from one form to another. In this case, the box has kinetic energy due to its motion, which will be converted into heat energy due to the friction force in the rough portion.

Using the equation for kinetic energy, KE=1/2mv^2, we can find the initial kinetic energy of the box before it enters the rough portion. Then, using the work-energy theorem, W=ΔKE, we can find the final kinetic energy of the box after it exits the rough portion. Finally, we can solve for the velocity using the equation KE=1/2mv^2.

c) To find the length of rough surface that will bring the box to a complete stop, we need to understand that the friction force acts in the opposite direction of the motion of the box, thus causing it to slow down. As we calculated in part (b), the friction force will convert the kinetic energy of the box into heat energy. Eventually, the box will come to a complete stop when all of its kinetic energy has been converted into heat energy.

Using the same principles as in part (b), we can set the final kinetic energy equal to zero and solve for the distance d using the equation W=Fd. This