Question

A 3.00 kg block starts from rest at the top of a 30° incline and accelerates uniformly down the incline, moving 1.83 m in 1.80 s.

(a) Find the magnitude of the acceleration of the block. m/s2

(b) Find the coefficient of kinetic friction between the block and the incline.

(c) Find the magnitude of the frictional force acting on the block. N

(d) Find the speed of the block after it has slid a distance 1.83 m. m/s

Answer #1

Mass of block = m = 3 kg

Angle of incline =
= 30^{o}

Initial speed of block = V_{1} = 0 m/s

Speed of block after sliding a distance 1.83m =
V_{2}

Distance slid by the block = d = 1.83 m

Time taken to slide through a distance of 1.83 m = t = 1.8 sec

Acceleration of block = a

Coefficient of kinetic friction =

d = V_{1}t + a(t_{1}^{2})/2

1.83 = (0)(1.8) + a(1.8^{2})/2

a = 1.13 m/s^{2}

V_{2} = V_{1} + at

V_{2} = 0 + (1.13)(1.8)

V_{2} = 2.034 m/s

From the free body diagram,

N = mgCos

f = N

f = mgCos

ma = mgSin- f

ma = mgSin - mgCos

a = gSin -gCos

1.13 = 9.81Sin(30) - (9.81)Cos(30)

= 0.444

f = mgCos

f = (0.444)(3)(9.81)Cos(30)

f = 11.31 N

a) Acceleration of the block = 1.13 m/s^{2}

b) Coefficient of kinetic friction between block and incline = 0.444

c) Friction force acting on block = 11.31 N

d) Speed of the block after is has slid down 1.83m = 2.034 m/s

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