In Figure (a), a 1.8 N force is applied to a 4.7 kg block at a downward angle th

Question

In Figure (a), a 1.8 N force is applied to a 4.7 kg block at a downward angle theta = 35degree as the block moves rightward through 1.4 m across a frictionless floor. Find the speed vf of the block at the end of that distance if the block's initial velocity is (a) 0 and (b) 1.2 m/s to the right, (c) The situation in Figure (b) is similar in that the block is initially moving at 1.2 m/s to the right, but now the 1.8 IM force is directed downward to the left. Find the speed v_f of the block at the end of the 1.4 m distance.

Explanation / Answer

1.8 N force
m= 4.7 kg
angle = 35 deg
d=1.4 m
a) initial velocity v1= 0

Work done W =F.d = F*cos(theta)*d

from work-energy theorem

W = change in kinetic energy

W = KE2-KE1

W = 0.5 m (v2^2-v1^2)

F*cos(theta)*d =  0.5 m (v2^2-v1^2)

1.8*cos(35)*1.4 = 0.5*4.7*(v2^2 - 0)

v2 = 0.937 m/s

b)initial velocity = 1.2 m/s

F*cos(theta)*d =  0.5 m (v2^2-v1^2)

1.8*cos(35)*1.4 = 0.5*4.7*(v2^2 - 1.2^2)

v2 = 1.523 m/s
c)initial velocity v1= 1.2 m/s
F=1.8 N
d= 1.4 m

-F*cos(theta)*d = 0.5 m*(v2^2-v1^2)

-1.8*cos(35)*1.4 = 0.5*4.7*(v2^2-1.2^2)

v2 = 0.75 m/s

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