A block of mass 10kg lies on a rough plane as shown in the figure. A constant force F is applied for 10 seconds, find out the value of work done by the force F.
A
12000J
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B
10000J
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C
−6000J
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D
0J (because friction is present)
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Solution
The correct option is A12000J FBD of Block
Net force in vertical direction N−mg=0 N=mg=10×10=100N
Frictional force =μN =0.2×100 =20N
Net force in horizontal direction =F−Frictional force ma=60−20 a=4010=4m/s2
Applying , s=ut+12at2=0+12×4×102 s=200m
Work done by force W=F.s=60×200 =12000J