Question

A box of mass m=10.0 kg, initially at rest, is pushed by a constant horizontal force of F=25.0 N for a distance d=10.0 m before reaching the bottom of a hill. The force stops and the box slides up the hill of height h=2.0 m. It hits a spring which compresses by an amount x=2.5 meters before the box stops.a) What is the speed of the box before it climbs up the hill?b) What is the spring constant?

Answer 1

a)

Since the force of 25 N is applied by a distance of 10 meters, we can calculate the work done:

`\begin{gathered} W=F\cdot d \\ W=25\cdot10 \\ W=250\text{ N} \end{gathered}`

Then, if all the energy is kinetic energy, we can calculate the box speed:

`\begin{gathered} E_k=(mv^2)/(2) \\ 250=(10\cdot v^2)/(2) \\ 5v^2=250 \\ v^2=50 \\ v=7.07\text{ m/s} \end{gathered}`

b)

The box slides up the hill, so part of the energy is now potential energy:

`\begin{gathered} E_p=m\cdot g\cdot h \\ E_p=10\cdot9.8\cdot2 \\ E_p=196\text{ J} \end{gathered}`

The remaining kinetic energy (250 - 196 = 54 J) will be converted into elastic potential energy of the spring:

`\begin{gathered} E_(ep)=(1)/(2)kx^2 \\ 54=(1)/(2)\cdot k\cdot2.5^2 \\ 6.25k=108 \\ k=17.28\text{ N/m} \end{gathered}`