A 4.60 kg block is set into motion up an inclined plane with an initial speed of

Question

A 4.60 kg block is set into motion up an inclined plane with an initial speed of v0 = 8.00 m/s (Fig. P8.33). The block comes to rest after traveling 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 frictional force exerted on the block (assumed to be constant).
N
(d) What is the coefficient of kinetic friction?

Explanation / Answer

here,

v0 = 8 m/s

d = 3 m

theta = 30 degree

a)

the change in block's kinetic energy , KE = 0.5 * m* v0^2

KE = 0.5 * 4.6 * 8^2

KE = 147.2 J

b)

the change in potential energy = m * g * ( d * sin(theta))

change in potential energy , PE = 4.6 * 9.8 * 3 * sin(30)

PE = 67.62 J

c)

let the friction force be F

work done by friction force = KE - PE

F * d = 147.2 - 67.62

F * 3 = 79.58

F = 26.53 N

the friction force is 26.53 N

d)

let the coefficient of friction be uk

uk * m * g * cos(theta) = F

uk * 4.6 * 9.8 * cos(30) = 26.53

uk = 0.679

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