A force is applied upon a body of 40 kg then its velocity increases from 1m/s to 2 m/s .Determine work done by the force.

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Solution

Given,

$M a s s = 40 k g, v = 1 m / s, v = 2 m / s$

According to the work-energy conversion theorem;

Work done by the force $=$ Change in Kinetic energy,

Work done by the force $= (\frac{1}{2}) \times m \times (v^{2} - u^{2})$

Work done $= 0.5 \times 40 \times (4 - 1) = 60 J$

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A force is applied upon a body of 40 kg then its velocity increases from 1m/s to 2 m/s .Determine work done by the force.

Given,Mass=40kg,v=1m/s,v=2m/sAccording to the work-energy conversion theorem;Work done by the force= Change in Kinetic energy,Work done by the force =(12)×m×(v2−u2)Work done =0.5×40×(4−1)=60J

Given,

$M a s s = 40 k g, v = 1 m / s, v = 2 m / s$

According to the work-energy conversion theorem;

Work done by the force $=$ Change in Kinetic energy,

Work done by the force $= (\frac{1}{2}) \times m \times (v^{2} - u^{2})$

Work done $= 0.5 \times 40 \times (4 - 1) = 60 J$