A wooden box slides down on a rough inclined plane. The angle of the incline is

Question

A wooden box slides down on a rough inclined plane. The angle of the incline is 30° above the horizontal. The weight of the object is 30 N. The velocity of the box is constant. Refer to the figure for problem1 1) What is the acceleration of the box? 2) 3) What is the magnitude of the kinetic friction force? 4) What is the coefficient of kinetic friction? 5) What is the net (total) work done on the box? a) 9.8 m/s2 What is the magnitude of the normal force? a) 30 N b) 0 m/s2 b) 15 N b) 20 N b) 0.3 b) 60 J 4.9 m/s c) 26 N c) 26 N d) 19.6 m/s d) 20 N d) 30 N a) 15 N c) 85 J

Explanation / Answer

here,

weight , W = 30 N

theta = 30 degree

a)

as it is moving with constant velocity

the accelration of the box is b) 0 m/s^2

b)

the magnitude of the normal force , N = W * cos(theta)

N = 30 * cos(30) = 26 N

c)

the magnitude of kinetic friction force , fk = the gravitational foce

fk = m * g * sin(theta)

fk = 15 N

d)

let the coefficient of kinetic friction be uk

uk * N = fk

uk * 26 = 15

uk = 0.6

e)

as the accelration is zero

the net force is zero

so the work done is also zero

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