(D) Given that the body starts from rest (implied by uniform acceleration starting from $t=0$),the acceleration $a$ is given by $a = \frac{v - 0}{t_1}

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$\frac{1}{2} m \frac{v^2}{t_1^2} t^2$

(D) Given that the body starts from rest (implied by uniform acceleration starting from $t=0$),the acceleration $a$ is given by $a = \frac{v - 0}{t_1} = \frac{v}{t_1}$.At any time $t$,the velocity $v_t$ of the body is $v_t = u + at = 0 + \left( \frac{v}{t_1} \right) t = \frac{v}{t_1} t$.According to the Work-Energy Theorem,the work done $W$ is equal to the change in kinetic energy:$W = \Delta K = \frac{1}{2} m v_t^2 - 0$.Substituting the value of $v_t$:$W = \frac{1}{2} m \left( \frac{v}{t_1} t \right)^2 = \frac{1}{2} m \frac{v^2}{t_1^2} t^2$.

Class 11 Physics Work, Energy, Power and Collision Work Energy Theorem and Conservation of Mechanical Energy

Difficult

$A$ body of mass $m$ is moving with uniform acceleration such that at time $t_1$ its velocity is $v$. What is the work done on the body at any time $t$?

A
$\frac{1}{2} m \frac{v}{t_1} t^2$
B
$m \frac{v}{t_1} t^2$
C
$\frac{1}{2} m \left( \frac{mv}{t_1} \right)^2 t^2$
D
$\frac{1}{2} m \frac{v^2}{t_1^2} t^2$

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Work, Energy, Power and Collision

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Work Energy Theorem and Conservation of Mechanical Energy

$A$ body of mass $m$ starts moving from rest along the $x-$axis such that its velocity varies as $v = a\sqrt{s}$,where $a$ is a constant and $s$ is the distance covered by the body. The total work done by all the forces acting on the body in the first $t$ seconds after the start of the motion is:

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$A$ body of mass $2 \, kg$ is projected vertically upwards with a kinetic energy of $490 \, J$. At what height will its kinetic energy become half of its initial value? $(g = 9.8 \, m/s^2)$

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The work-energy theorem is valid in the presence of:

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$A$ body of mass $2 \, kg$ slides down a curved track which is a quadrant of a circle of radius $1 \, m$ as shown in the figure. All the surfaces are frictionless. If the body starts from rest,its speed at the bottom of the track is ............. $m/s$.

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$A$ body of mass $m$ is moving with uniform acceleration such that at time $t_1$ its velocity is $v$. What is the work done on the body at any time $t$?

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$A$ body of mass $m$ starts moving from rest along the $x-$axis such that its velocity varies as $v = a\sqrt{s}$,where $a$ is a constant and $s$ is the distance covered by the body. The total work done by all the forces acting on the body in the first $t$ seconds after the start of the motion is:

$A$ body of mass $2 \, kg$ is projected vertically upwards with a kinetic energy of $490 \, J$. At what height will its kinetic energy become half of its initial value? $(g = 9.8 \, m/s^2)$

Potential energy of a body of mass $1 \ kg$ free to move along $X$-axis is given by $U(x) = (\frac{x^2}{2} - x) \ J$. If the total mechanical energy of the body is $2 \ J$,then the maximum speed of the body is (Assume only conservative force acts on the body).

The work-energy theorem is valid in the presence of:

$A$ body of mass $2 \, kg$ slides down a curved track which is a quadrant of a circle of radius $1 \, m$ as shown in the figure. All the surfaces are frictionless. If the body starts from rest,its speed at the bottom of the track is ............. $m/s$.

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