$A$ block of mass $m$ moving with a velocity $v_0$ on a smooth horizontal surface strikes and compresses a spring of stiffness $k$ until the mass comes to rest,as shown in the figure. This phenomenon is observed by two observers:
$A$: standing on the horizontal surface
$B$: standing on the block
To an observer $A$,the work done by the normal reaction $N$ between the block and the spring on the block is

$A$ force acts on a body of mass $10 \ kg$,resulting in its displacement given as $x = \left(\frac{t^3}{25}\right) \ m$,where $t$ is the time in seconds. The work done by the force in the first $2 \ s$ is (in $J$)

$A$ particle moves under the effect of a force $F = cx$ from $x = 0$ to $x = x_1$. The work done in the process is

$A$ $1 \,kg$ box placed at the origin starts sliding along the $x$-axis under the action of a force $\vec{F} = F \hat{i}$. Its acceleration as a function of $x$ is given by $a(x) = \beta x$,where $\beta = 5 \,s^{-2}$. The work done by $\vec{F}$ in moving the box from $x = 2 \,cm$ to $x = 5 \,cm$ in joules is:

$A$ body of mass $3 \ kg$ moves under the influence of a force such that its displacement is given by $s = t^3/3 \ m$. What is the work done by the force in the first $2 \ s$ in $J$?

An object of mass '$m$' initially at rest on a smooth horizontal plane starts moving under the action of a force $F = 2 \text{ N}$. In the process of its linear motion,the angle $\theta$ (as shown in the figure) between the direction of the force and the horizontal varies as $\theta = kx$,where $k$ is a constant and $x$ is the distance covered by the object from its initial position. The expression for the kinetic energy of the object is $E = \frac{n}{k} \sin \theta$. The value of $n$ is .....