Work Power Energy Questions

A particle of mass $100\text{g}$ is thrown vertically upwards with a speed of $5\text{m/s}$. The work done by the force of gravity during the time the particle goes up is:

A ball of mass $0.2\text{kg}$ is thrown vertically upwards by applying a force by hand. If the hand moves $0.2\text{m}$ while applying the force and the ball goes up to $2\text{m}$ height further, find the magnitude of the force. (Consider $g=10{\text{m/s}}^2$)

The potential energy of a $1\text{kg}$ particle free to move along the $x$ -axis is given by $V(x)=\left(\frac{x^4}{4}-\frac{x^2}{2}\right)\text{J}$. The total mechanical energy of the particle is $2\text{J}$. Then, the maximum speed (in m/s) is:

A $2\text{kg}$ block slides on a horizontal floor with a speed of $4\text{m/s}$. It strikes an uncompressed spring and compresses it till the block is motionless. The kinetic friction force is $15\text{N}$ and the spring constant is $10,000\text{N/m}$. The spring compresses by:

The upper half of an inclined plane with inclination $\varphi$ is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by:

A bullet fired into a fixed target loses half of its velocity after penetrating $3\text{ cm}$. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?

A spherical ball of mass $20\text{ kg}$ is stationary at the top of a hill of height $100\text{ m}$. It rolls down a smooth surface to the ground, then climbs up another hill of height $30\text{ m}$ and finally rolls down to a horizontal base at a height of $20\text{ m}$ above the ground. The velocity attained by the ball is:

The block of mass $M$ moving on the frictionless horizontal surface collides with a spring of spring constant $K$ and compresses it by length $L$. The maximum momentum of the block after collision is:

A body of mass $m$ is accelerated uniformly from rest to a speed $v$ in a time $T$. The instantaneous power delivered to the body as a function time is given by:

If two springs $S_1$ and $S_2$ of force constants $k_1$ and $k_2$, respectively, are stretched by the same force, it is found that more work is done on spring $S_1$ than on spring $S_2$.Statement 1: If stretched by the same amount, work done on $S_1$, will be more than that on $S_2$.Statement 2: $k_1<k_2$.