Work Energy and Power MCQ Questions & Answers in Basic Physics | Physics

$$F = - \frac{{dU}}{{dy}} = b - 2ay$$

When the block $$B$$ is displaced towards wall 1, only spring $${S_1}$$ is compressed and $${S_2}$$ is in its natural state. This happens because the other end of $${S_2}$$ is not attached to the wall but is free. Therefore the energy stored in the system $$ = \frac{1}{2}{k_1}{x^2}.$$   When the block is released, it will come back to the equilibrium position, gain momentum, overshoot to equilibrium position and move towards wall 2. As this happens, the spring $${S_1}$$  comes to its natural length and $${S_2}$$ gets compressed. As there are no frictional forces involved, the P.E. stored in the spring $${S_1}$$ gets stored as the P.E. of spring $${S_2}$$ when the block $$B$$ reaches its extreme position after compressing $${S_2}$$ by $$y.$$
$$\eqalign{ & \therefore \frac{1}{2}{k_1}{x^2} = \frac{1}{2}{k_2}{y^2} \cr & \frac{1}{2} \times k{x^2} = \frac{1}{2}4k{y^2},\,\,\,\,{x^2} = 4{y^2} \cr & \therefore \frac{y}{x} = \frac{1}{2} \cr} $$

$$m = 10 \times 0.8\,kg = 8\,kg$$
height of iron chain $$= 5m$$
$$P = \frac{{mgh}}{t} = \frac{{8 \times 10 \times 5}}{{10}}W = 40\,W$$

$$P = 10\;kW = 10000\;W$$
Fuel consumption $$= 1\,g/s$$
Calorific value $$= 2\,kcal/g$$
$$\therefore $$ Energy produced $$= 2\,kcal/s$$
Input power $$= 2\,kcal/s = 2000\,cal/s$$
$$ = 2000 \times 4.18\,J/s = 8.4\,kW$$
$$\therefore $$ This claim is invalid.

Applying conservation of linear momentum, we write, $${m_1}{u_1} = {m_2}{u_2}$$
$$\eqalign{ & {\text{Here,}}\,{m_1} = 18\,kg,\,{m_2} = 12\,kg \cr & {u_1} = 6\,m{s^{ - 1}},{u_2} = ? \cr & \therefore 18 \times 6 = {u_2} \cr & \Rightarrow {u_2} = \frac{{18 \times 6}}{{12}} = 9\,m{s^{ - 1}} \cr} $$
Thus, kinetic energy of $$12\,kg$$  mass
$$\eqalign{ & {K_2} = \frac{1}{2}{m_2}u_2^2 \cr & = \frac{1}{2} \times 12 \times {\left( 9 \right)^2} \cr & = 6 \times 81 = 486\;J \cr} $$

The potential energy of a system increases, if work is done by the system against a conservative force.
$$ - \Delta U = {W_{{\text{conservative force}}}}$$

$$\eqalign{ & \frac{1}{2}M{v^2} = \frac{1}{2}k{L^2} \cr & \Rightarrow v = \sqrt {\frac{k}{M}} .L \cr & {\text{Momentum}} = M \times v = M \times \sqrt {\frac{k}{M}} .L \cr & = \sqrt {kM} .L \cr} $$

$${v_{0y}} = {v_0}\sin \theta ,$$   this component does not change
$${v_{0x}} = {v_0}\cos \theta \,{\text{and}}\,{u_x}\cos \theta $$
For elastic collision relative velocity before collision is equal to relative velocity after collision along $$x$$ -axis
$$\eqalign{ & {v_0}\cos \theta - u\cos \theta = u\cos \theta + {v_x} \cr & {\text{if}}\,{v_x}\cos \theta = {v_{0y}} = \sin \theta \cr} $$
By solving $$u = 0.1$$

$$\eqalign{ & {\text{Initial,}}\,K.E. = \frac{1}{2}m{v^2} = \frac{1}{2} \times \frac{{20}}{{1000}} \times 600 \times 600 = 3600\,J \cr & {\text{Change}}\,{\text{in}}\,\,K.E. = P.E. \cr & \frac{1}{2}m\left( {{v^2} - {{v'}^2}} \right) = mgh \cr & \Rightarrow 3600 - \frac{1}{2} \times \frac{{20}}{{1000}} \times v_1^2 = 4 \times 10 \times 80 \cr & \Rightarrow {v_1} = 200\,m/s \cr} $$

Let the blow compress the spring by $$x$$ before stopping.
Kinetic energy of the block = ($$P.E$$  of compressed spring) + work done against function.
$$\eqalign{ & \frac{1}{2} \times 2 \times {\left( 4 \right)^2} = \frac{1}{2} \times 10,000 \times {x^2} + \left( { + 15} \right) \times x \cr & 10,000{x^2} + 30x - 32 = 0 \cr & \Rightarrow 5000{x^2} + 15x - 16 = 0 \cr & \therefore x = - \frac{{15 \pm \sqrt {{{\left( {15} \right)}^2} - 4 \times \left( {5000} \right)\left( { - 16} \right)} }}{{2 \times 5000}} \cr & = 0.055\,m = 5.5\,cm \cr} $$