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?

Jan 29, 2017

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

Explanation:

This problem is most easily done by conservation of energy.
Three energy forms are involved:
A change in kinetic energy: $\frac{1}{2} m {v}_{f}^{2} - \frac{1}{2} m {v}_{i}^{2}$

A change in gravitational potential energy: $m g h$

Frictional heating: $\mu {F}_{N} \Delta d$

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 $\Delta d$ (its displacement along the ramp).

$h = \Delta \mathrm{ds} \int h \eta$

Also, we must note that the normal force on an incline is not equal to mg, but to $m g \cos \theta$, where $\theta = \frac{\pi}{4}$ in this case. With all that looked after, our equation becomes

$\frac{1}{2} m {v}_{f}^{2} - \frac{1}{2} m {v}_{i}^{2} + m g \Delta \mathrm{ds} \in \left(\frac{\pi}{4}\right) + \mu m g \cos \left(\frac{\pi}{4}\right) \Delta d = 0$

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:

$- \frac{1}{2} {\left(2\right)}^{2} + \left(9.8\right) \Delta d \left(0.707\right) + \left(\frac{8}{7}\right) \left(9.8\right) \left(0.707\right) \Delta d = 0$

$- 2.0 + 6.93 \Delta d + 7.92 \Delta d = 0$

$14.85 \Delta d = 2.0$

$\Delta d = 0.135 m$