Two objects having masses in the ratio $1:4$ are at rest. When both of them are subjected to the same force separately,they achieve the same kinetic energy during times $t_1$ and $t_2$ respectively. Then the ratio of $t_1$ to $t_2$ is

$A$ body is moved from rest along a straight line by a machine delivering constant power. The ratio of displacement and velocity $(s/v)$ varies with time $t$ as :

$A$ body of mass $1\,kg$ is thrown upwards with a velocity $20\,m/s$. It momentarily comes to rest after attaining a height of $18\,m$. How much energy is lost due to air friction? (Given $g = 10\,m/s^2$)

$A$ constant force $\vec{F} = 3\hat{i} - 2\hat{j} - \hat{k} \text{ N}$ causes a displacement $\vec{r} = 2\hat{i} - 3\hat{j} - 3\hat{k} \text{ m}$ in $2 \text{ s}$. The work done and the power are respectively:

The mass of a person is $60 \, kg$. If he gains $10^5 \, \text{calories}$ of heat energy from food and the efficiency of his body is $28 \%$,then up to what height can he climb? (approximately) ($g = 9.8 \, m/s^2$,$J = 4.2 \, J/\text{cal}$)

$A$ block of mass $m$ starts from rest and slides down a frictionless semi-circular track from a height $h$ as shown. When it reaches the lowest point of the track,it collides with a stationary piece of putty also having mass $m$. If the block and the putty stick together and continue to slide,the maximum height that the block-putty system could reach is: