1.

A body acted upon by a variable force F = 3 +0.5.x. Work done in moving the body from x=0 to x = 4m is:

Answer»

8 J
32 J
24 J
16 J

Solution :Work done, `W=int_(0)^(4)(3+0.5 X)DX`
`=3int_(0)^(4)dx+0.5int_(0)^(4)XDX`
`=3[x]_(0)^(4)+0.5[(X^(2))/(2)]_(0)^(4)=3(4-0)+0.5[(16)/(2)-0]`
=16 J


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