Asked by Sierra
The 2 kg block in Figure 7-25 slides down a frictionless curved ramp, starting from rest at a height of h = 3 m. The block then slides d = 12 m on a rough horizontal surface before coming to rest.
Figure 7-25
(a) What is the speed of the block at the bottom of the ramp?
(b) What is the energy dissipated by friction?
(c) What is the coefficient of friction between the block and the horizontal surface?
Figure 7-25
(a) What is the speed of the block at the bottom of the ramp?
(b) What is the energy dissipated by friction?
(c) What is the coefficient of friction between the block and the horizontal surface?
Answers
Answered by
Damon
a)
(1/2) m v^2 = m g h
so
v = sqrt(2 g h)
b)
m g h
c)
mu m g * 12 = m g h
so
mu = h/12 = 3/12 = .25
(1/2) m v^2 = m g h
so
v = sqrt(2 g h)
b)
m g h
c)
mu m g * 12 = m g h
so
mu = h/12 = 3/12 = .25
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