Question

A 1.2-kg block, travelling at 14.2-m/s, encounters a ramp with a coefficient of kinetic friction of 0.15. The ramp is tilted 31° above the horizontal. Use work and energy arguments to answer the following.

How far along the ramp does the block travel?

Assume the coefficient of static friction is very small and the block slides back down the ramp. What is its speed at the bottom?

Answer 1

Given,

m = 1.2 kg ; v = 14.2 m/s ; uk = 0.15 ; theta = 31 deg

from conservation of energy

KE = PE + Wf

1/2 m v^2 = m g h + uk m g d cos(theta)

0.5 v^2 = g d sin(theta) + uk g d cos(theta)

d = 0.5 v^2/ [g sin(theta) + uk g cos(theta)]

d = 0.5 x 14.2^2/ 9.81 (sin31 + 0.15 x cos31) = 15.97 m

Hence, d = 15.97 m

PE at top = m g d sin(theta)

from conservation of energy

1/2 m v'^2 = m g d sin(theta)

v' = sqrt (2 g d sin(theta))

v' = sqrt (2 x 9.81 x 15.97 x sin31) = 12.7 m/s

Hence, v' = 12.7 m/s