A) 9.82 kJ
B) 89 Kj
C) 4.35 kJ
D) 5.17 kJ
Correct Answer: D
Solution :
Here : Weight of block \[w=2kN,\] Distance d = 10 m Angle of inclination on the plane \[\alpha =15{}^\circ \] The block will be pulled up on a smooth plane Hence, force of resistance due to inclination \[F=\omega \,\,\sin \alpha =2\times {{10}^{3}}\sin 15{}^\circ \] \[=2\times {{10}^{3}}\times 0.2588\] \[=0.5176\,kN\] Now work done, \[W=Fd=0.5176\times {{10}^{3}}\times 10\] \[=5.17\,kN\]You need to login to perform this action.
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