A body of mass 2 kg moves under the influence of a force- Its position x changes with time according to the relation x-t3-3 where x is in meter and t in seconds- The work done by this force in first two seconds will be

x = t^3/3 , v = dx/dt = t^2 w = KE, v_1 = 0, v_2 = 4 w = 1/2mv_2^2 1/2mv_1^2 w = 1/2× 2 [(4)^2 (0)^2] = 16 ...

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Standard XII

Physics

Relation between Work Done and PE

A body of mas...

Question

A body of mass 2kg moves under the influence of a force. Its position x changes with time according to the relation $x = \frac{t^{3}}{3}$ where x is in meter and t in seconds. The work done by this force in first two seconds will be

1600 Joule

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16 Joule

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1.6 Joule

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Solution

The correct option is C 16 Joule
$x = \frac{t^{3}}{3}$ , $v = \frac{d x}{d t} = t^{2}$
$w = △ K E$ , $v_{1} = 0$ , $v_{2} = 4$
$w = \frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2}$
$w = \frac{1}{2} \times 2 [(4)^{2} - (0)^{2}] = 16$

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A body of mass 2kg moves under the influence of a force. Its position x changes with time according to the relation x=t33 where x is in meter and t in seconds. The work done by this force in first two seconds will be

The correct option is C 16 Joule x=t33 , v=dxdt=t2 w=△KE, v1=0, v2=4 w=12mv22−12mv21 w=12×2[(4)2−(0)2]=16

A body of mass 2kg moves under the influence of a force. Its position x changes with time according to the relation $x = \frac{t^{3}}{3}$ where x is in meter and t in seconds. The work done by this force in first two seconds will be

The correct option is C 16 Joule
$x = \frac{t^{3}}{3}$ , $v = \frac{d x}{d t} = t^{2}$
$w = △ K E$ , $v_{1} = 0$ , $v_{2} = 4$
$w = \frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2}$
$w = \frac{1}{2} \times 2 [(4)^{2} - (0)^{2}] = 16$