Question

A box with an initial speed of `2 m/s` is moving up a ramp. The ramp has a kinetic friction coefficient of `8/7` and an incline of `( pi )/4`. How far along the ramp will the box go?

Answer 1

The box will travel `0.135m` along the ramp.

Explanation:

This problem is most easily done by conservation of energy.
Three energy forms are involved:
A change in kinetic energy: `1/2mv_f^2-1/2mv_i^2`

A change in gravitational potential energy: `mgh`

Frictional heating: `muF_NDelta d`

Before we can continue, we need to be aware of a couple of complications

We must express `h` (the height the object rises) in terms of `Deltad` (its displacement along the ramp).

`h=Deltadsintheta`

Also, we must note that the normal force on an incline is not equal to mg, but to `mgcostheta`, where `theta=pi/4` in this case.

With all that looked after, our equation becomes

`1/2mv_f^2-1/2mv_i^2+mgDeltadsin(pi/4)+mumgcos(pi/4)Delta d=0`

Notice that we can divide every term by the mass m, (including the right side of the equation)

Also, `v_f=0` , and inserting the values for `mu`, `v_i` and g, we get:

`-1/2(2)^2+(9.8)Deltad(0.707)+(8/7)(9.8)(0.707)Delta d=0`

`-2.0+6.93Deltad+7.92Deltad=0`

`14.85Deltad=2.0`

`Deltad=0.135m`