An object of mass $20 \,kg$ is displaced by $x = 5t^2 \,m$ (where $t$ is time) due to the application of a force. The ratio of the work done in $3 \,s$ and $5 \,s$ is:

Under the action of a force,a $2 \, kg$ body moves such that its position $x$ as a function of time $t$ is given by $x = \frac{t^2}{3}$,where $x$ is in metres and $t$ is in seconds. The work done by the force in the first two seconds is .......... $J$.

$A$ simple pendulum is released from point $A$ as shown in the figure. If $m$ and $l$ represent the mass of the bob and the length of the pendulum respectively,the gain in kinetic energy at point $B$ is:

$A$ particle $(m = 1 \; kg)$ slides down a frictionless track $(AOC)$ starting from rest at a point $A$ (height $2 \; m$). After reaching $C$,the particle continues to move freely in air as a projectile. When it reaches its highest point $P$ (height $1 \; m$),the kinetic energy of the particle (in $J$) is: (Figure drawn is schematic and not to scale; take $g = 10 \; ms^{-2}$)

As shown in the figure,a particle of mass $10 \ kg$ is placed at a point $A$. When the particle is slightly displaced to its right,it starts moving and reaches the point $B$. The speed of the particle at $B$ is $x \ m/s$. (Take $g = 10 \ m/s^2$). The value of $x$ to the nearest integer is:

$A$ student of mass $M$ is $1.5 \,m$ tall and has her centre of mass $1 \,m$ above the ground when standing straight. She wants to jump up vertically. To do so,she bends her knees so that her centre of mass is lowered by $0.2 \,m$ and then pushes the ground by a constant force $F$. As a result,she jumps up such that the maximum height of her feet is $0.3 \,m$ above the ground. The ratio $F / Mg$ is (in $.5$)