(A) Given, mass of particle, $m = 30 \,g = 3 \times 10^{-2} \,kg$.The position of the particle is $x = \alpha t^2$.Velocity $v = \frac{dx}{dt} = 2\alp

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(A) Given, mass of particle, $m = 30 \,g = 3 \times 10^{-2} \,kg$.The position of the particle is $x = \alpha t^2$.Velocity $v = \frac{dx}{dt} = 2\alpha t$.Acceleration $a = \frac{dv}{dt} = 2\alpha$.Force $F = ma = (3 \times 10^{-2} \,kg) \times (2 \times 1 \,m/s^2) = 6 \times 10^{-2} \,N$.Work done $W = \int F dx$. Since $x = \alpha t^2$, $dx = 2\alpha t dt$.At $t = 0, x = 0$. At $t = 4, x = 1 \times (4)^2 = 16 \,m$.$W = \int_{0}^{16} (6 \times 10^{-2}) dx = 6 \times 10^{-2} \times [x]_{0}^{16} = 6 \times 10^{-2} \times 16 = 0.96 \,J$.

Class 11 Physics Work, Energy, Power and Collision Work Done by Constant Force

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$A$ 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 $x = \alpha t^2$, where $x$ is in metre, $t$ is in seconds and $\alpha = 1 \,m/s^2$. The work done during the first $4 \,s$ is (in $J$)

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$0.96$
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$0.45$
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$0.49$
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$0.53$

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Work, Energy, Power and Collision

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Work Done by Constant Force

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$A$ 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 $x = \alpha t^2$, where $x$ is in metre, $t$ is in seconds and $\alpha = 1 \,m/s^2$. The work done during the first $4 \,s$ is (in $J$)

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$A$ force $F = (5\hat{i} + 3\hat{j}) \ N$ acts on an object,causing a displacement of $r = (2\hat{i} - 1\hat{j}) \ m$. Calculate the work done in $J$.

The kinetic energy $(KE)$ acquired by a mass $m$ in travelling a certain distance $d$,starting from rest,under the action of a constant force $F$ is ..........

$A$ force $\overrightarrow{F} = 5\hat{i} + 6\hat{j} + 4\hat{k}$ acting on a body produces a displacement $\overrightarrow{S} = 6\hat{i} - 5\hat{k}$. The work done by the force is: (in $J$)

$A$ force $\vec{F} = 3\hat{i} + c\hat{j} + 2\hat{k}$ acting on a particle causes a displacement $\vec{s} = -4\hat{i} + 2\hat{j} + 3\hat{k}$. If the work done is $6\,J$,then the value of $c$ is:

$A$ cyclist comes to a skidding stop in $10 \; m$. During this process,the force on the cycle due to the road is $200 \; N$ and is directly opposed to the motion.$(a)$ How much work does the road do on the cycle?$(b)$ How much work does the cycle do on the road?

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$A$ cyclist comes to a skidding stop in $10 \; m$. During this process,the force on the cycle due to the road is $200 \; N$ and is directly opposed to the motion.
$(a)$ How much work does the road do on the cycle?
$(b)$ How much work does the cycle do on the road?