A 29 kg box is being pushed across the floor by a constant f

A 29 kg box is being pushed across the floor by a constant force ‹ 109, 0, 0 › N. The coefficient of kinetic friction for the table and box is 0.15. At t = 6.0 s the box is at location ‹ 11, 3, 3 › m, traveling with velocity ‹ 6, 0, 0 › m/s. What is its position and velocity at t = 7.5 s?

Solution

We have a block of 29 kg being pushed by a force of 109 N along the x axis. Also, the box has a velocity of 6 m/s along x axis and is located at x = 11 m.

As the box suffers no force or has any velocity along y and z axis, we can solve the problem for x axis individually and that will be our required answer.

Here, the net force suffered by the box will be given as 109 - friction

Or Fnet = 109 - 29*9.81*0.15 = 66.3265 N

or, the net acceleration = 66.3265 / 29 = 2.2871 m/s^2

Now, for the motion towards x axis we have:

S = ut + 0.5at^2 = 6(1.5) + 0.5*2.2871*1.5*1.5

or S = 11.57299 m

Hence the position of the box at t = 7.5 seconds would be 11+ 11.57299 = 22.57299 metres along x axis

Hence the position is <22.57299 , 3, -3>