Question

Figure 1: Block on incline. 2, A block of mass m = 5 kg is subject to a force of magnitude 20 N that makes an angle of p20 with the (frictionless) inclined plane of contact. The plane itself is inclined θ = 300 with the horizontal. (a) What is the acceleration of the block along the incline? (b) If the block begins at rest, in what direction will it move subsequently (i.e., up or down the incline)? (c) If there were friction with coefficient μ8-02, then would the block still slide if initially at rest?

Answer 1

(a)

net force along the incline

Fnet = m*g*sintheta - Fp*cosphi

Fnet = (5*9.8*sin30) - (20*cos20) = 5.706 N

from newton's second law Fnet = m*a

a = Fnet/m = 5.706/5 = 1.14 m/s^2

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(b)

along the incline m*g*sintheta > Fp*cos20

the block move down

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(c)

perpendicular to the incline

Fp*sinphi + N = m*g*costheta

normal force N = m*g*costheta - Fp*sinphi

N = (5*9.8*cos30) - (20*sin20) = 36 N

frictional force fs = us*N = 0.3*36 = 7.2 N

net force due to gravitiaon and Fp = m*g*sintheta - Fp*cosphi

Fnet = 5*9.8*sin30 - 20*cos20 = 5.71 N < fs

the block will not slide