1/2 m v^2 = m g h - friction work
50 v^2 = (100 * 10 * 50) - (100 * 150)
A flat 100kg rock slides down a smooth rocky hill. It started from rest. The hill is 100m long and 50m high. The friction force between the rock and the ground is 150N. Use g=10m/s².
Calculate the speed of the rock at the bottom of the hill
2 answers
potential at top = m g h = 100 * 10 *50 = 50,000 Joules
work done by friction, force opposite to motion so negative = -150 *100
= -15,000 Joules
so
kinetic energy left at bottom = (1/2)m v^2 = 35,000 Joules
50 v^2 = 35,000
v^2 = 3500/5 = 700
work done by friction, force opposite to motion so negative = -150 *100
= -15,000 Joules
so
kinetic energy left at bottom = (1/2)m v^2 = 35,000 Joules
50 v^2 = 35,000
v^2 = 3500/5 = 700
1 answer