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

$${U_1} = mg{h_1}\,{\text{and}}\,{U_2} = mg{h_2}$$
$$\% $$ energy lost $$ = \frac{{{U_1} - {U_2}}}{{{U_1}}} \times 100$$
$$\eqalign{ & = \frac{{mg{h_1} - mg{h_2}}}{{mg{h_1}}} \times 100 \cr & = \left( {\frac{{{h_1} - {h_2}}}{{{h_1}}}} \right) \times 100 \cr & = \frac{{2 - 1.5}}{2} \times 100 \cr & = 25\% \cr} $$

Work done by a variable force $$F$$ in displacement from $$x = {x_1}$$   to $$x = {x_2}$$  is given by
$$\eqalign{ & W = \int_{{x_1}}^{{x_2}} F \left( {dx} \right) \cr & {\text{Given,}}\,\,{x_1} = 0,{x_2} = 5 \cr & F = \left( {7 - 2x + 3{x^2}} \right)N \cr & \therefore W = \int_0^5 {\left( {7 - 2x + 3{x^2}} \right)} dx \cr & = \left[ {7x - {x^2} + {x^3}} \right]_0^5 \cr & = \left[ {7 \times 5 - {{\left( 5 \right)}^2} + {{\left( 5 \right)}^3}} \right] \cr & = \left[ {35 - 25 + 125} \right] \cr & = 135\,J \cr} $$

When stone is at its lowest position, it has only kinetic energy, given by $$K = \frac{1}{2}m{u^2}$$

At the horizontal position, it has energy $$E = U + K = \frac{1}{2}m{{u'}^2} + mgl$$
According to conservation of mechanical energy, $$K = E$$

$$\eqalign{ & \therefore \frac{1}{2}m{u^2} = \frac{1}{2}m{{u'}^2} + mgl \cr & {\text{or}}\,\,\frac{1}{2}m{{u'}^2} = \frac{1}{2}m{u^2} - mgl \cr & {\text{or}}\,\,{{u'}^2} = {u^2} - 2gl \cr & {\text{or}}\,\,u' = \sqrt {{u^2} - 2gl} \,......\left( {\text{i}} \right) \cr} $$
So, the magnitude of change in velocity
$$\eqalign{ & \left| {\Delta u} \right| = \left| {u' - u} \right| \cr & = \sqrt {{{u'}^2} + {u^2} + 2u'u\cos {{90}^ \circ }} \cr} $$

$$\eqalign{ & \left| {\Delta u} \right| = \sqrt {{{u'}^2} + {u^2}} \cr & = \sqrt {2\left( {{u^2} - gl} \right)} \,\,\left[ {{\text{from}}\,{\text{Eq}}{\text{.}}\left( {\text{i}} \right)} \right] \cr} $$

$$\eqalign{ & {\text{Given,}}\,\frac{{dk}}{{dt}} = {\text{constant}} \cr & \Rightarrow k \propto t \Rightarrow v \propto \sqrt t \cr & {\text{Also,}}\,P = Fv = \frac{{dk}}{{dt}} = {\text{constant}} \cr & \Rightarrow F \propto \frac{1}{v} \Rightarrow F \propto \frac{1}{{\sqrt t }} \cr} $$

According to principle of conservation of energy
Potential energy = kinetic energy
$$ \Rightarrow mgh = \frac{1}{2}m{v^2} \Rightarrow v = \sqrt {2gh} $$
If $${h_1}$$ and $${h_2}$$ are initial and final heights, then
$$ \Rightarrow {v_1} = \sqrt {2g{h_1}} ,{v_2} = \sqrt {2g{h_2}} $$
Loss in velocity,
$$\Delta v = {v_1} - {v_2} = \sqrt {2g{h_1}} - \sqrt {2g{h_2}} $$
$$\therefore $$ fractional loss in velocity
$$\eqalign{ & = \frac{{\Delta v}}{{{v_1}}} = \frac{{\sqrt {2g{h_1}} - \sqrt {2g{h_2}} }}{{\sqrt {2g{h_1}} }} = 1 - \sqrt {\frac{{{h_2}}}{{\;{h_1}}}} \cr & = 1 - \sqrt {\frac{{1.8}}{5}} = 1 - \sqrt {0.36} = 1 - 0.6 = 0.4 = \frac{2}{5} \cr} $$

$$\eqalign{ & U = \left( {\frac{1}{2}} \right)M{v^2} \cr & U \times \frac{2}{{{v^2}}} = M\,{\text{or}}\,M = \frac{{2U}}{{{v^2}}} \cr} $$

Energy lost in the collision is
$$\Delta E = \frac{1}{2}mu_{rel}^2\left( {1 - {e^2}} \right)$$
$$\therefore \frac{3}{{16}}{\left( {{{\overrightarrow u }_1} - {{\overrightarrow u }_2}} \right)^2} = \frac{1}{2}mu_{rel}^2\left( {1 - {e^2}} \right) \Rightarrow e = 0.5$$

$$\eqalign{ & E = Pt = 45 \times {10^3} \times 50 \times 60 = 1.35 \times {10^8} \cr & V = \frac{{1.35 \times {{10}^8}}}{{3.6 \times {{10}^7}}} = 3.75\,L \cr} $$
Since the motor is $$20\% $$  efficient, five times as much gasoline is needed.

$$\eqalign{ & W = \int\limits_{{x_1}}^{{x_2}} {Fdx} = \int\limits_4^{ - 2} { - 6{x^3}dx} \cr & = - 6\left| {\frac{{{x^4}}}{4}} \right|_4^{ - 2} = \frac{{ - 6}}{4}\left[ {{{\left( { - 2} \right)}^4} - {{\left( 4 \right)}^2}} \right] = 360\,J. \cr} $$

$$\eqalign{ & Pt = nmgh \cr & 2400\left( {60} \right) = n\left( {65} \right)\left( {9.8} \right)\left( 6 \right);n = 37 \cr} $$