Question

An 8.70-kg block slides with an initial speed of 1.80 m/s down a ramp inclined at an angle of 25.3 ∘ with the horizontal. The coefficient of kinetic friction between the block and the ramp is 0.87. Use energy conservation to find the distance the block slides before coming to rest.

Answer 1

The initial kinetic energy of the block is Ekin = 1/2 mv^2
The work done by the friction is u*mg*cos25.3*s
The work done by lifting the box up by h is mgh and h is s*sin25.3 --> so this work is Epot = mg*s*sin25.3.

The initial kinetic energy will be the sum of the friction work and the potential energy:

1/2mv^2 = 0.62*mg*cos25.3*s + mg*sin25,3*s
1/2 v^2 = s*g(0,62*cos25.3 + sin25.3)
s = 1/2v^2/(g(0,62*cos25.3 + sin25.3))

s = 1/2(1.80)^2/(g(0,62*cos25.3 + sin25.3))

s = 0.1673