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Question

A constant force F1=103N applied at an angle 30 with horizontal on body (m=5 kg) at rest for 8 second. In this time interval, body moves from position A to B. To move from position B to A, another constant force F2=20 N is applied on body . If work done by this force is 960 J, then find the angle at which the force is acting on the body with horizontal.


A
30
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B
45
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C
60
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D
90
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Solution

The correct option is C 60
When F1=103 N is applied,
Fx=F1cosθ=103×cos30=15 N
ax=Fx/m=153=5 m/s2
S=ut+12at2(u=0)

Distance travelled by body from A to B is
S=12×3×82=96 m

Let F2=20 N acts on body makes an angle θ with horizontal
W=F2Scosθ
960=20×96cosθ
cosθ=12
θ=60


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