Work, Energy and Power · Physics · TS EAMCET

A constant force of $(8 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) \mathrm{N}$ acting on a body of mass 2 kg displaces the body from $(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}) \mathrm{m}$ to $(4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) \mathrm{m}$. The work done in the process is

The kinetic energy of a body of mass 4 kg moving with a velocity of $(2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \mathrm{ms}^{-1}$ is

A block kept on a frictionless horizontal surface is connected to one end of a horizontal spring of constant $100 \mathrm{Nm}^{-1}$ whose other end is fixed to a rigid vertical wall. Initially the block is at its equilibrium position. The block is pulled to a distance of 8 cm and released.The kinetic energy of the block when it is a distance of 3 cm from the mean position is

The work done in blowing a soap bubble of volume $V$ is $W$. The work done in blowing the bubble of volume 2 V from the same soap solution is

A body thrown vertically upwards from the ground reaches a maximum height $h$. The ratio of the kinetic and potential energies of the body at a height $40 \%$ of $h$ from the ground is

A block of mass $m$ with an initial kinetic energy $E$ moves up an inclined plane of inclination $\theta$. If $\mu$ is the coefficient of friction between the plane and the body, the work done against friction before coming to rest is

A man of mass 80 kg goes to the market on a scooter of mass 100 kg with certain speed. On application of brakes, the stopping distance is $s_1$. The man returns home on the same scooter, with the same speed with a 60 kg bag of rice. If $s_2$ is the new stopping distance when the brakes are applied with the same force, then

A block of mass 0.5 kg is at rest on a horizontal table. The coefficient of kinetic friction between the table and the block is 0.2 . If a horizontal force of 5 N is applied on the block, the kinetic energy of the block in a time of 4 s is (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

The work done in increasing the diameter of a soap bubble from 2 cm to 4 cm is (Surface tension of soap solution $=3.5 \times 10^{-2} \mathrm{Nm}^{-1}$ )