A particle of mass 0-01 kg travels along a curve with velocity given by 4-16 k- ms-1- After some time- its velocity becomes 8-20 - ms-1 due to the action of a conservative force- The work done on particle during this interval of time is

V_1=√(4^2+16^2)=√(272) and v_2=√(8^2+20^2)=√(464) Work done = Change in kinetic energy =1/2m[v_2^2 v_1^2]=1/2× 0.01[464 272]=0.96J. ...

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Standard XII

Physics

Work Done as Dot Product

A particle of...

Question

A particle of mass 0.01 kg travels along a curve with velocity given by $4^i + 16^k m s^{- 1}$ . After some time, its velocity becomes $8^i + 20^j m s^{- 1}$ due to the action of a conservative force. The work done on particle during this interval of time is

0.32 J

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6.9 J

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9.6 J

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0.96 J

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Solution

The correct option is D
0.96 J

$v_{1} = \sqrt{4^{2} + 16^{2}} = \sqrt{272} a n d v_{2} = \sqrt{8^{2} + 20^{2}} = \sqrt{464}$
Work done = Change in kinetic energy = $\frac{1}{2} m [v_{2}^{2} - v_{1}^{2}] = \frac{1}{2} \times 0.01 [464 - 272] = 0.96 J .$

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A particle of mass 0.01 kg travels along a curve with velocity given by 4^i+16^k ms−1. After some time, its velocity becomes 8^i+20^j ms−1 due to the action of a conservative force. The work done on particle during this interval of time is

The correct option is D 0.96 J v1=√42+162=√272 and v2=√82+202=√464 Work done = Change in kinetic energy =12m[v22−v21]=12×0.01[464−272]=0.96J.

A particle of mass 0.01 kg travels along a curve with velocity given by $4^i + 16^k m s^{- 1}$ . After some time, its velocity becomes $8^i + 20^j m s^{- 1}$ due to the action of a conservative force. The work done on particle during this interval of time is

The correct option is D
0.96 J

$v_{1} = \sqrt{4^{2} + 16^{2}} = \sqrt{272} a n d v_{2} = \sqrt{8^{2} + 20^{2}} = \sqrt{464}$
Work done = Change in kinetic energy = $\frac{1}{2} m [v_{2}^{2} - v_{1}^{2}] = \frac{1}{2} \times 0.01 [464 - 272] = 0.96 J .$