A 5.20–kg block is set into motion up an inclined plane with an initial speed of
ID: 1364795 • Letter: A
Question
A 5.20–kg block is set into motion up an inclined plane with an initial speed of vi = 7.40 m/s (see figure below). The block comes to rest after traveling d = 3.00 m along the plane, which is inclined at an angle of ? = 30.0° to the horizontal.
(a) For this motion, determine the change in the block's kinetic energy. J
(b) For this motion, determine the change in potential energy of the block–Earth system. J
(c) Determine the friction force exerted on the block (assumed to be constant). N
(d) What is the coefficient of kinetic friction?
Explanation / Answer
a) delta_KE = KE2 - KE1
= 0.5*m*(v2^2 - v1^2)
= 0.5*5.2*(0^2 - 7.4^2)
= -142.376 J
b) change in potentail energy = m*g*d*sin(theta)
= 5.2*9.8*3*sin(30)
= 76.44 J
c) Workdone by friction = change in mechanical energy
= -142.376 + 76.44
= -65.936 J
Frictional force = W_friction/d
= 65.936/3
= 21.98 N
d) W_friction = mue_k*N*d*cos(180)
W_friction = - mue_k*m*g*cos(30)*d
==> mue_k = -W_friction/(m*g*cos(30)*d)
= -(-65.936)/(5.2*9.8*cos(30)*3)
= 0.498
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