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Standard XII

Physics

Relation between Work Done and PE

A force acts ...

Question

A force acts on a 30 gm particle in such a way that the position of the particle as a function of time is given by $x = 3 t - 4 t^{2} + t^{3}$ , where x is in metres and t is in seconds. The work done during the first 4 seconds is

5.28 J

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450 mJ

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490 mJ

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530 mJ

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Solution

The correct option is A
5.28 J

$v = \frac{d x}{d t} = 3 - 8 t + 3 t^{2}$

$∴ v_{0} = 3 m / s a n d v_{4} = 19 m / s$

$W = \frac{1}{2} m (v_{4}^{2} - v_{0}^{2})$ (According to work energy theorem)

$= \frac{1}{2} \times 0.03 \times (19^{2} - 3^{2}) = 5.28 J$

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Similar questions

A force acts on a 30 gm particle in such a way that the position of the particle as a function of time is given by $x = 3 t - 4 t^{2} + t^{3}$ , where x is in metres and t is in seconds. The work done during the first 4 seconds is

Q. A force acts on a

30 g m

particle in such a way that the position of the particle as a function of time is given by

x = 3 t - 4 t^{2} + t^{3}

, where

x

is in metres and

t

is in seconds. The work done on the particle during the first

4

second is

Q. A force acts on a

30

g particle in such a way that the position of the particle as a function of time is given by

x = 3 t - 4 t^{2} + t^{3}

, where x is in metre and t is in second. The work done during the first

4

second is?

Q. A force acts on a

30 g

particle in such a way that the position of the particle as a function of time is given by

x = 3 t - 4 t^{2} + t^{3}

, where

x

is in meters and

t

is in seconds. The work done during the first

4

seconds is

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A force acts on a 30 gm particle in such a way that the position of the particle as a function of time is given by x=3t−4t2+t3, where x is in metres and t is in seconds. The work done during the first 4 seconds is

The correct option is A 5.28 J v=dxdt=3−8t+3t2 ∴v0=3m/s and v4=19m/s W=12m(v24−v20) (According to work energy theorem) =12×0.03×(192−32)=5.28 J

A force acts on a 30 gm particle in such a way that the position of the particle as a function of time is given by $x = 3 t - 4 t^{2} + t^{3}$ , where x is in metres and t is in seconds. The work done during the first 4 seconds is

The correct option is A
5.28 J

$v = \frac{d x}{d t} = 3 - 8 t + 3 t^{2}$

$∴ v_{0} = 3 m / s a n d v_{4} = 19 m / s$

$W = \frac{1}{2} m (v_{4}^{2} - v_{0}^{2})$ (According to work energy theorem)

$= \frac{1}{2} \times 0.03 \times (19^{2} - 3^{2}) = 5.28 J$