A block of mass 25 kg is pulled along a horizontal surface by a force at an angle 45 with the horizontal. The friction coefficient between the block and the surface is 0.25. The block travels at a uniform velocity. The work done by the applied force during a displacement of 5 m of the block is :
- 490 J
- 970 J
- 735 J
- 245 J
Answer
(4) 245 J
Solution
Forces on the block (taking g = 9.8 m s-2):
[figure]
Vertically the forces balance, N + F 45 = mg, so
N = mg - F 45
The velocity is uniform, so horizontally F 45 = f = N = (mg - F 45 ).
Since 45 = 45 , this gives
F 45 (1 + ) = mg
F 45 = 0.25 25 9.81.25 = 49 N
Hence the work done by the applied force is
W = F 45 s = 49 5
W = 245 J
Answer: (4)