h = 9.7*sin30 = 4.85 m.
V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*4.85 = 95.06
V = 9.75 m/s.
Starting from rest, a(n)9 kg block slides 9.7 m
down a frictionless ramp (inclined at 30◦
from
the floor) to the bottom. The block then
slides an additional 23.5 m along the floor
before coming to stop.
The acceleration of gravity is 9.8 m/s
2.Find the speed of the block at the bottom
of the ramp.
Answer in units of m/s
1 answer
3 answers