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 \[F=w\text{ }sin\text{ }\theta \] \[=mg\text{ }sin\text{ }{{15}^{o}}\] \[=2sin\,{{15}^{o}}\] \[=0.5176\text{ }kN\] Also, work done = force \[\times \] displacement \[=0.5176\times 10=\text{ }5.17\text{ }kJ\]You need to login to perform this action.
You will be redirected in
3 sec