Question

Draw a free body diagram for the block showing the
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Answer #1

a)

b)

along y direction


F*sin30 + mg = N

c)

frictional force = f = u*N = u*(F*sin30 + mg)

d)


along x direction

F*cos30-f = m*a


e)

F*cos30 - u*(F*sin30 + mg) = m*a


a = [ F*cos30 - u*(F*sin30 + mg) ]/ m

f)


a = ( (1500*cos30)-0.3*((1500*sin30)+(220*9.8)) )/220

a = 0.132 m/s^2

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