A) 9.82 kJ
B) 8.91 kJ
C) 5.17 kJ
D) 4.36 kJ
Correct Answer: C
Solution :
The various forces acting on a block kept on inclined plane are as shown. The force of resistance due to inclination when block is pulled upwards is \[F=w\sin \theta \] \[=mg\sin {{15}^{o}}\] \[=2\sin {{15}^{o}}\] = 0.5176 kN Also, work done = force x displacement \[=0.571\times 10=5.17\,kJ\]You need to login to perform this action.
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