Sliding Box?

Question

A 1.50 kg box slides down a rough incline plane from a height h of 1.03 m. The box had a speed of 2.77 m/s at the top and a speed of 2.34 m/s at the bottom. Calculate the mechanical energy lost due to friction (as heat, etc.).

Answer 1

The energy at the top should be equal to the energy at the bottom, IF there was no energy loss. Thus to solve for the energy loss, first calculate the energy at the top:

Energy at the top =Potential energy (PE)plus Kinetic energy (KE)
= mgh+1/2mv^2

where g=9.8m/s^2, h=1.03m, and v=2.77m/s

Substitute given values:

Energy at top =m*9.8*1.03+!/2m*2.77^2
=10.09m+3.84m
=13.93m

Then compute for the energy at the bottom:

Energy at the bottom=PE at bottom +KE at bottom

where h=0 and v=2.34m/s

Substitute given values:

Energy at bottom =mgh+1/2mv^2
=m*9.8*0+1/2m*2.34^2
=2.74m

Energy loss=Energy at the top - Energy at the bottom
=13.93m-2.74m
=11.19m

Where m=1.50kg

Substitute value of m:
Energy loss= 11.19*1.50
=16.78 joules