Let X be that sliding distance.
Potential energy loss = work done against friction on the horizontal surface
M*g*(5.0 m) = M*g*(0.20)*X
M*g cancels out, and
X = 25 m
A block slides from rest with negligible friction down a track, descending 5.0m to point P at the bottom. It then slides on the horizontal surface. The coefficient of friction between the block and the horizontal surface is 0.20. How far does the block slide on the horizontal surface before it comes to rest?
2 answers
Time taken for block to reach the bottom, just as reaches horizontal surface is 't'
s = ut+1/2 at2
5 = 1/2*9.8*t2
t = 1.01 sec
Final velocity as it reaches start of horizontal surface is v1 = gt = 9.8*1.01 = 9.899 m/s
Distance it travels on horizontal surface before coming to rest is s = ut+0.5*at^2
a = - 0.2 g
t = u/a = 9.899/(0.2*9.8)
t = 5.05 seconds
s = 9.899(5.05) -0.5*(0.2*9.8*5.05^2)
s = 24.99 m
s = ut+1/2 at2
5 = 1/2*9.8*t2
t = 1.01 sec
Final velocity as it reaches start of horizontal surface is v1 = gt = 9.8*1.01 = 9.899 m/s
Distance it travels on horizontal surface before coming to rest is s = ut+0.5*at^2
a = - 0.2 g
t = u/a = 9.899/(0.2*9.8)
t = 5.05 seconds
s = 9.899(5.05) -0.5*(0.2*9.8*5.05^2)
s = 24.99 m
1 answer