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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Work, Power and Energy

A force of 0.5 N acts on body of mass 5 kg initially at rest for one minute. Find the work done by the force.

\[(a)\;60\;J \quad (b)\;70\;J \quad (c)\;180\;J \quad (d)\;90\;J\]

1 Answer

Answer: 90 J
Work is done by a force on a body over a certain displacement. $W = F\; s$
In this case $F = 0.5 N$. We need to calculate displacement and then Work.
Given $F = 0.5N$ and mass $m = 5Kg \rightarrow$ Acceleration $a = \large\frac{F}{m}$$ = \large\frac{0.5}{5}$$ = 0.1\;m/s^2$
Given $t = 1m = 60s$ and Initial velocity $u = 0\;m/s$, we can calculate Displacement $s = ut + \large\frac{1}{2}$$ut^2$$ = \large\frac{1}{2}$$ (0.1\; m/s^2)(60^2 s^2)$$ = 180m$
Hence work done $ = F\;s = 0.5N \times 180m = 90J$
answered Jul 17, 2013 by meena.p
edited Aug 25, 2014 by balaji.thirumalai
 

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