Under the action of a force- a 1 kg body moves- such that its position x as a function of time t is given by x-t32 where x is in meter and t is in second- The work done by the force in first 3 second is

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Standard X

Physics

Force

Under the act...

Question

Under the action of a force, a $1 k g$ body moves, such that its position $x$ as a function of time $t$ is given by $x = \frac{t^{3}}{2}$ where $x$ is in meter and $t$ is in second. The work done by the force in first $3$ second is

243 J

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\frac{729}{8}

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24.3 J

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Solution

The correct option is C $\frac{729}{8}$
We can get acceleration by double differentiating x wrt t ,
So we get a = 3t which gives F=3t N

Now Work done = $\int F d x = \int \frac{3 t^{3}}{2} d t = \frac{9 t^{4}}{8} = \frac{729}{8}$

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Under the action of a force, a 1 kg body moves, such that its position x as a function of time t is given by x=t32 where x is in meter and t is in second. The work done by the force in first 3 second is

The correct option is C 7298We can get acceleration by double differentiating x wrt t ,So we get a = 3t which gives F=3t NNow Work done = ∫Fdx=∫3t32dt=9t48=7298

Under the action of a force, a $1 k g$ body moves, such that its position $x$ as a function of time $t$ is given by $x = \frac{t^{3}}{2}$ where $x$ is in meter and $t$ is in second. The work done by the force in first $3$ second is

The correct option is C $\frac{729}{8}$
We can get acceleration by double differentiating x wrt t ,
So we get a = 3t which gives F=3t N

Now Work done = $\int F d x = \int \frac{3 t^{3}}{2} d t = \frac{9 t^{4}}{8} = \frac{729}{8}$