Question
A 3 kg block travels along a horizontal surface with a coefficient of kinetic friction of 0.26 at a speed of 7 m/s. After sliding a distance of 1.0 m the block makes a smooth transition to a ramp with a coefficient of kinetic friction of 0.26. How far up the ramp does the block travel before coming to a momentary stop.
Answers
initial ke = (1/2)mv^2 =.5*3*49
= 73.5 Joules
slides for one meter, how much energy lost?
mg = 3* 9.81 = 29.4 Newtons weight
so
mu m g = friction force = .26*29.4 = 7.65 N
so friction work = 7.65 * 1 = 7.65 Joules lost to friction
so
ke at bottom of ramp = 73.5 - 7.65 = 65.8 J
say ramp slope angle from horizontal = T
friction force down slope = 7.65 cos T
weight component down slope
= mg sin T=29.4 sin T
so
work done moving x meters up slope
= (7.65 cos T + 29.4 sin T) x = 65.8 Joules
= 73.5 Joules
slides for one meter, how much energy lost?
mg = 3* 9.81 = 29.4 Newtons weight
so
mu m g = friction force = .26*29.4 = 7.65 N
so friction work = 7.65 * 1 = 7.65 Joules lost to friction
so
ke at bottom of ramp = 73.5 - 7.65 = 65.8 J
say ramp slope angle from horizontal = T
friction force down slope = 7.65 cos T
weight component down slope
= mg sin T=29.4 sin T
so
work done moving x meters up slope
= (7.65 cos T + 29.4 sin T) x = 65.8 Joules
This problem is just like the one Bob Pursley answered for you. Please try yourself!