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

The above situation can also be looked upon as the decrease in the gravitational potential energy of spring mass system is equal to the gain in spring elastic potential energy.
$$Mgx = \frac{1}{2}k{r^2},\,\,\,x = \frac{{2Mg}}{k}$$

Velocity of free falling body does not depend on its mass, it depends on the height from which it has been dropped.
$${v_1} = {v_2} = v$$   at a $$30\,ft$$  height from falling point.
Here, $${m_1} = 2\,kg,\,\,{m_2} = 4\,kg$$
Thus, $$\frac{{{K_1}}}{{{K_2}}} = \frac{{\frac{1}{2}{m_1}{v^2}}}{{\frac{1}{2}{m_2}{v^2}}} = \frac{{{m_1}}}{{{m_2}}} = \frac{2}{4} = \frac{1}{2}$$

According to work-energy theorem, work done by a force in displacing a body measures the change in kinetic energy of the body or work done by a force is equal to change in $$KE$$  of the body.
or $$W = \Delta E$$
$$\eqalign{ & = {\text{final }}KE - {\text{initial }}KE \cr & = \frac{1}{2}m{v^2} - \frac{1}{2}m{u^2} \cr} $$
Given, $$v = 1000\,m/s,\,m = 10\,g$$
$$u = 500\,m/s,\, = 0.01\,kg$$
Putting the values of $${m_1}{u_1}v$$
$$\eqalign{ & \therefore W = \frac{1}{2} \times 0.01\left[ {{{\left( {1000} \right)}^2} - {{\left( {500} \right)}^2}} \right] \cr & = 3750\,J \cr} $$

$$\eqalign{ & {\text{We}}\,{\text{knowthat}}\,\,F \times v = {\text{Power}} \cr & \therefore F \times v = c\,\,{\text{where}}\,c = {\text{constant}} \cr & \therefore m\frac{{dv}}{{dt}} \times v = c\,\,\left( {\therefore F = ma = \frac{{mdv}}{{dt}}} \right) \cr & \therefore m\int\limits_0^v v dv = c\int\limits_0^t d t\,\,\therefore \frac{1}{2}m{v^2} = ct \cr & \therefore v = \sqrt {\frac{{2c}}{m}} \times {t^{\frac{1}{2}}} \cr & \therefore \frac{{dx}}{{dt}} = \sqrt {\frac{{2c}}{m}} \times {t^{\frac{1}{2}}}\,\,{\text{where}}\,v = \frac{{dx}}{{dt}} \cr & \int\limits_0^x {dx} = \sqrt {\frac{{2c}}{m}} \times \int\limits_0^t {{t^{\frac{1}{2}}}dt} \cr & x = \sqrt {\frac{{2c}}{m}} \times \frac{{2{t^{\frac{3}{2}}}}}{3} \Rightarrow x \propto {t^{\frac{3}{2}}} \cr} $$

Total initial energy $$ = \frac{1}{2}{m_1}u_1^2 + \frac{1}{2}{m_2}u_2^2$$
Since, after collision one particle absorb energy $$\varepsilon .$$
∴ Total final energy $$ = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2 + \varepsilon $$
From conservation of energy,
$$\eqalign{ & \frac{1}{2}{m_1}u_1^2 + \frac{1}{2}{m_2}u_2^2 = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2 + \varepsilon \cr & \Rightarrow \frac{1}{2}{m_1}u_1^2 + \frac{1}{2}{m_2}u_2^2 - \varepsilon = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2 \cr} $$

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

$$W.D.$$  from ground frame $$ = 6.4 \times 50 = 320\,J$$
From elevator frame, displacement $$= 0.$$
$$N\cos {37^ \circ } = 8 \times 0.8 = 6.4\,N$$

$$10 \times 0.8 = 8\,N$$

$$\eqalign{ & W = {\text{area of }}F - x{\text{ graph}} \cr & = {\text{area of }}\Delta + {\text{area of rectangle}} + {\text{area of }}\Delta \cr & = \frac{{5 \times 3}}{2} + 10 \times 3 + \frac{{5 \times 3}}{2} = 45\,J \cr} $$

From conservation of linear momentum,
Initial momentum $${p_{{\text{initial}}}} =$$  Find momentum $${p_{{\text{final}}}}$$
$$\eqalign{ & 0 = {m_1}{v_1} - {m_2}{v_2} \cr & {\text{or}}\,\,{m_1}{v_1} = {m_2}{v_2} \cr & {\text{or}}\,\,\frac{{{v_1}}}{{{v_2}}} = \frac{{{m_1}}}{{{m_2}}}\,......\left( {\text{i}} \right) \cr} $$
Thus, ratio of kinetic energies,
$$\frac{{{K_1}}}{{{K_2}}} = \frac{{\frac{1}{2}{m_1}v_1^2}}{{\frac{1}{2}{m_2}v_2^2}} = \frac{{{m_1}}}{{{m_2}}} \times {\left( {\frac{{{m_2}}}{{{m_1}}}} \right)^2} = \frac{{{m_2}}}{{{m_1}}}$$
NOTE
In a collision of two bodies, whether it is perfectly elastic or inelastic, linear momentum is always conserved but kinetic energy need not be conserved.

Area under $$F-t$$   graph gives the impulse or the change in the linear momentum of the body. As the initial velocity (and therefore the initial linear momentum) of the body is zero, the area under $$F-t$$   graph gives the final linear momentum of the body.

Area of $$\Delta \,AOB$$
$$\eqalign{ & = \frac{1}{2} \times 3 \times 4 = 6N - s \cr & {\text{Also }}\frac{{OA}}{{OB}} = \frac{{CD}}{{CB}} \cr & \Rightarrow \frac{4}{3} = \frac{{CD}}{{1.5}}f \cr & \Rightarrow CD = 2 \cr} $$
$$\therefore $$ Area of $$\Delta BCD$$
$$\eqalign{ & = - \left[ {\frac{1}{2} \times 1.5 \times 2} \right] \cr & = - 1.5\,N - s \cr} $$
$$\therefore $$ The final linear momentum $$=6-1.5=4.5 \,N-s$$
$$\therefore $$ Kinetic energy of the block
$$ = \frac{{{p^2}}}{{2m}} = \frac{{{{\left( {4.5} \right)}^2}}}{{2 \times 2}} = 5.06\,J$$