The 2 kg block in the figure slides down a frictionless curved ramp, starting fr
ID: 1281545 • Letter: T
Question
The 2 kg block in the figure slides down a frictionless curved ramp, starting from rest at a height of h = 3 m. The block then slides d = 7 m on a rough horizontal surface before coming to rest.What is the coefficient of friction between the block and the horizontal surface?
The 2 kg block in the figure slides down a frictionless curved ramp, starting from rest at a height of h = 3 m. The block then slides d = 7 m on a rough horizontal surface before coming to rest.What is the coefficient of friction between the block and the horizontal surface?Explanation / Answer
WHEN IT FALLS TO GROUND LEVEL :
mgh = 1/2 mv2 ( from energy conservation )
V =SQRT (2*g*h )
V =SQRT (2*10*5 )
v=10 m/s
now as the horizontal level is of frictin nature
ACCORDING TO WORK ENERGY THEOREM:
WORK DONE BY ALL THe FORCES = CHANGE IN K.E
- f * s = (1/2)mvf2 - (1/2)mvi2
Vf is zero AS FINAL VELOCITY IS ZERO
(umg ) *s = (1/2) m (10)2
u*m*10 *9 = (1/2) *m*10*10
u=0.555
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