Potential energy at the top of the hill
= mgh
= 10*9.81*5.8 J
=569 J
Heat energy produced
= 414 J
If we assume conservation of energies, energy at the bottom of the hill in the form of kinetic energy
=(1/2)mv² = 569+414 =983
Solve for v:
v=sqrt(2*983/10)
=14.0 m/s
A 10 kg block slides up a hill to a height of 5.8 m. If 414 J of thermal energy are generated, how fast was the block going at the bottom of the hill?
1 m/s
1 answer