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 $4$ second is?

5.28

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450

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490

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530

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Solution

The correct option is A $5.28$ J
$M = 20 g m = 0.020 k g$

$x = 3 t - 4 t ² + t ³ m$

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

$a = \frac{d v}{d t} = - 8 + 6 t \frac{m}{s^{2}}$

Force $= m * a = 0.120 t - 0.160 N$

Workdone $= d W = F d x = F * \frac{d x}{d t} * d t = (0.120 t - 0.160) * (3 - 8 t + 3 t ²) d t = 0.360 t ³ - 1.440 t ² + 1.640 t - 0.480$

$W = 0.090 t ⁴ - 0.490 t ³ + 0.82 t ² - 0.480 t = 2.88 J$

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

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

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 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

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A force acts on a 30gm 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 second is?

The correct option is A 5.28JM=20gm=0.020kgx=3t−4t²+t³mv=dxdt=3−8t+3t²msa=dvdt=−8+6tms2Force=m∗a=0.120t−0.160NWorkdone=dW=Fdx=F∗dxdt∗dt=(0.120t−0.160)∗(3−8t+3t²)dt=0.360t³−1.440t²+1.640t−0.480W=0.090t⁴−0.490t³+0.82t²−0.480t=2.88J

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 $4$ second is?