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Physics
Work, Energy and Power
71 questions, each with its answer.
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The power of a crane, which lifts a mass of 1000 kg to a height of 20 m in 10 s is: ($g = 9.8\ m/s^2$)Power of crane $= \frac{mgh}{t} = \frac{1000 \times 9.8 \times 20}{10}$ $= 19.6 \times 1000$ W = 19.6 kW
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A particle of mass M moves along a horizontal x axis from x = 0 to x = L. The coefficient of kinetic friction varies as a function of x as $\mu_k(x)=\mu_0-\alpha x$, where $\mu_0$, $\alpha$ are constants of appropriate dimensions, so that $\mu_k(L)=0$. The total work done by the frictional force during the motion is $n\mu_0MgL$, where g is the acceleration due to gravity. The value of n is:$\mu_k=\mu_0-\alpha x$ and $\mu_k(L)=0 \Rightarrow \alpha=\frac{\mu_0}{L}$. Work done by kinetic friction $=-\int_0^L(\mu_0-\alpha x)mg\,dx=-\left[\mu_0mgL-\alpha mg\frac{L^2}{2}\right]=-\left[\mu_0mgL-\frac{\mu_0}{L}mg\frac{L^2}{2}\right]=-\frac{\mu_0mgL}{2}$. Comparing the magnitude with $n\mu_0mgL$: $n=\frac{1}{2}$.
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Bob B of mass m at rest is hanging vertically from the ceiling via a massless string of length 10 m, as shown in the figure. Point mass A of mass m travelling horizontally with speed 10 ms$^{-1}$ hits bob B elastically. The bob B rises h meter after the collision. Taking the acceleration due to gravity g = 10 ms$^{-2}$ and neglecting the size of the bob, the value of h is:
The masses are equal and the collision is head-on and elastic, so the velocities interchange: B moves off with 10 m/s. By conservation of mechanical energy, $\frac{1}{2}mv^2=mgh \Rightarrow h=\frac{v^2}{2g}=\frac{100}{20}=5$ m. -
The kinetic energies of two similar cars A and B are $100\text{ J}$ and $225\text{ J}$ respectively. On applying brakes, car A stops after $1000\text{ m}$ and car B stops after $1500\text{ m}$. If $F_A$ and $F_B$ are the forces applied by the brakes on cars A and B, respectively, then the ratio $F_A / F_B$ is:
From work-energy theorem, $F = \frac{KE}{s}$. Therefore, $F_A / F_B = \frac{100/1000}{225/1500} = \frac{0.10}{0.15} = \frac{2}{3}$.
According to the Work-Energy Theorem, work done by the braking force equals the change in kinetic energy: $W = F \cdot s = \Delta KE \implies F = \frac{KE}{s}$. For car A: $F_A = \frac{100\text{ J}}{1000\text{ m}} = 0.10\text{ N}$. For car B: $F_B = \frac{225\text{ J}}{1500\text{ m}} = 0.15\text{ N}$. The ratio is $\frac{F_A}{F_B} = \frac{0.10}{0.15} = \frac{2}{3}$. -
A bob of heavy mass $m$ is suspended by a light string of length $l$. The bob is given a horizontal velocity $v_0$ as shown in figure. If the string gets slack at some point P making an angle $\theta$ from the horizontal, the ratio of the speed $v$ of the bob at point P to its initial speed $v_0$ is:
Setting tension $T = 0$ at slack point P gives $v = \sqrt{gl \sin\theta}$. Using conservation of energy gives $v_0 = \sqrt{gl(2 + 3\sin\theta)}$, so $v/v_0 = \sqrt{\frac{\sin\theta}{2 + 3\sin\theta}}$.
At point P where the string goes slack, tension $T = 0$. The radial component of gravity provides the necessary centripetal force: $mg \sin\theta = \frac{m v^2}{l} \implies v^2 = gl \sin\theta$. By conservation of mechanical energy between the lowest point and point P: $\frac{1}{2} m v_0^2 = \frac{1}{2} m v^2 + mg(l + l \sin\theta)$. Multiplying by $2/m$: $v_0^2 = v^2 + 2gl(1 + \sin\theta) = gl \sin\theta + 2gl + 2gl \sin\theta = gl(2 + 3\sin\theta)$. Thus, the ratio of speeds is $\frac{v}{v_0} = \sqrt{\frac{gl \sin\theta}{gl(2 + 3\sin\theta)}} = \left(\frac{\sin\theta}{2 + 3\sin\theta}\right)^{1/2}$. -
At any instant of time $t$, the displacement of any particle is given by $2t-1$ (SI unit) under the influence of force of 5N. The value of instantaneous power is (in SI unit):Instantaneous velocity $v = \frac{ds}{dt} = \frac{d}{dt}(2t - 1) = 2\ ms^{-1}$ Instantaneous power $P = Fv = 5 \times 2 = 10$ W
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Two bodies A and B of same mass undergo completely inelastic one dimensional collision. The body A moves with velocity $v_1$ while body B is at rest before collision. The velocity of the system after collision is $v_2$. The ratio $v_1 : v_2$ is;Applying conservation of momentum: $mv_1 + 0 = 2mv_2$ $\Rightarrow v_2 = \frac{v_1}{2}$ $\frac{v_1}{v_2} = \frac{v_1}{v_1/2} = 2$ $v_1 : v_2 = 2 : 1$
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An object moving along horizontal x-direction with kinetic energy 10 J is displaced through $x=(3\hat{i})$ m by the force $\vec{F}=(-2\hat{i}+3\hat{j})$ N. The kinetic energy of the object at the end of the displacement x isBy the work-energy theorem, $W=\Delta KE$. $W=\vec{F}\cdot\vec{x}=(-2\hat{i}+3\hat{j})\cdot(3\hat{i})=-6$ J. So $K_f-10=-6\Rightarrow K_f=4$ J.
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An object falls from a height of 10 m above the ground. After striking the ground it loses 50% of its kinetic energy. The height upto which the object can rebounce from the ground is:KE before striking ground: $K_1=mgh_1=mg(10)$. After losing 50%, $K_2=\dfrac{1}{2}K_1=mg(5)$. Since $K_2=mgh_2$, $h_2=5$ m.
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The potential energy of a long spring when stretched by 2 cm is $U$. If the spring is stretched by 8 cm, potential energy stored in it will be:
Spring PE is $\frac{1}{2}kx^2$, so 4 times the stretch gives $16U$.
Potential energy stored in the spring $=\frac{1}{2}kx^2$ Now $\frac{1}{2}k(2)^2=U$ and $\frac{1}{2}k(8)^2=U'$ (say) $\Rightarrow U'=\frac{64}{4}U=16U$ -
A bullet from a gun is fired on a rectangular wooden block with velocity $u$. When bullet travels $24\ cm$ through the block along its length horizontally, velocity of bullet becomes $\frac{u}{3}$. Then it further penetrates into the block in the same direction before coming to rest exactly at the other end of the block. The total length of the block is:
Work-energy theorem for both stages gives total length $d=24\times\frac{9}{8}=27\ cm$.
$\frac{1}{2}m\left(\frac{u}{3}\right)^2-\frac{1}{2}mu^2=-F_R\times24$ $0-\frac{1}{2}mu^2=-F_R\times d$ $\frac{\frac{1}{2}mu^2}{\frac{1}{2}mu^2\times\frac{8}{9}}=\frac{d}{24}$ $d=24\times\frac{9}{8}=27\ cm$ -
A bullet of mass m hits a block of mass M elastically. The transfer of energy is the maximum, when:In a one-dimensional elastic collision, the maximum energy transfer from the bullet to the block occurs when the masses are equal, i.e. $M=m$.
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A particle moves with a velocity $(5\hat{i}-3\hat{j}+6\hat{k})$ m s$^{-1}$ horizontally under the action of constant force $(10\hat{i}+10\hat{j}+20\hat{k})$N. The instantaneous power supplied to the particle is:$P=\vec F\cdot\vec V=(10\hat i+10\hat j+20\hat k)\cdot(5\hat i-3\hat j+6\hat k)=50-30+120=140$ W.
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An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of $1.5\ ms^{-1}$. The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is : ($g=10\ ms^{-2}$)The lift moves with constant velocity, so a = 0. Tension (force by motor) $T = W + f = 2000 \times 10 + 3000 = 23000$ N Power $= Tv = 23000 \times 1.5 = 34500$ W
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The energy that will be ideally radiated by a 100 kW transmitter in 1 hour is :$E = P \times t = 100 \times 10^3 \times 3600$ $= 36 \times 10^7$ J
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Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional force are 10% of the input energy. How much power is generated by the turbine? ($g=10\ m/s^2$)$P_{in} = \frac{mgh}{t} = 15 \times 10 \times 60 = 9000$ W Losses are 10%, so $P_{out} = 90\%$ of $P_{in} = 0.9 \times 9000 = 8100$ W = 8.1 kW
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A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is ($g=10\ m/s^2$) nearly:Velocity just before striking the ground: $v_1 = \sqrt{2gh} = \sqrt{2 \times 10 \times 10} = 10\sqrt{2}$ m/s (downward) It rebounds to the same height, so the speed just after is the same, $10\sqrt{2}$ m/s (upward). Impulse $= m|\Delta v| = 0.15 \times [10\sqrt{2} - (-10\sqrt{2})] = 0.15 \times 20\sqrt{2}$ $= 3\sqrt{2} \approx 4.2$ kg m/s
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The energy equivalent of 0.5 g of a substance is:$E = mc^2 = 0.5 \times 10^{-3} \times (3 \times 10^8)^2$ $= 4.5 \times 10^{13}$ J
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Body A of mass $4m$ moving with speed $u$ collides with another body B of mass $2m$, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is:Fractional loss of KE of the colliding body in a head-on elastic collision: $\frac{\Delta KE}{KE}=\frac{4m_1m_2}{(m_1+m_2)^2}$ $=\frac{4\times4m\times2m}{(4m+2m)^2}=\frac{32m^2}{36m^2}=\frac{8}{9}$
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A force $F=20+10y$ acts on a particle in $y$-direction where $F$ is in newton and $y$ in meter. Work done by this force to move the particle from $y=0$ to $y=1$ m is:Work done by a variable force: $W=\int_{y_i}^{y_f}F\,dy$, with $y_i=0$, $y_f=1\ m$ $W=\int_0^1(20+10y)\,dy=\left[20y+\frac{10y^2}{2}\right]_0^1$ $W=20+5=25\ J$
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A particle of mass 5m at rest suddenly breaks on its own into three fragments. Two fragments of mass m each move along mutually perpendicular direction with speed v each. The energy released during the process is:Momentum conservation: $mv\hat{i}+mv\hat{j}+3m\vec{v}'=0 \Rightarrow |\vec{v}'|=\frac{\sqrt{2}v}{3}$. Energy released $=\frac{1}{2}mv^2+\frac{1}{2}mv^2+\frac{1}{2}(3m)\frac{2v^2}{9}=\frac{4}{3}mv^2$.
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An object of mass 500 g, initially at rest, is acted upon by a variable force whose X component varies with X in the manner shown. The velocities of the object at point X = 8 m and X = 12 m, would be the respective values of (nearly)
Work done = area under the F-x graph. Up to x = 8 m: $W=100+30=130$ J $=\Delta KE$, so $130=\frac{1}{2}\times\frac{1}{2}v^2 \Rightarrow v\approx22.8\approx23$ m/s. Similarly up to x = 12 m, $v\approx20.6$ m/s. -
A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to
Minimum speed at the bottom is $\sqrt{5gR}$, so $h = 5R/2 = 5D/4$.
As the track is frictionless, total mechanical energy remains constant.
$T.M.E_I = T.M.E_F \Rightarrow 0 + mgh = \frac{1}{2}mv_L^2 + 0$
$h = \frac{v_L^2}{2g}$
For completing the vertical circle, $v_L = \sqrt{5gR}$
$h = \frac{5gR}{2g} = \frac{5}{2}R = \frac{5}{4}D$ -
A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be
Momentum conservation gives $v/4$ for the heavy block; $e = (v/4)/v$.
According to the law of conservation of linear momentum, $mv + 4m\times0 = 4mv' + 0 \Rightarrow v' = \frac{v}{4}$ $e = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}} = \frac{v/4}{v}$ $e = \frac{1}{4} = 0.25$ -
Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m/s. Take g constant with a value 10 m/$s^2$. The work done by the (i) gravitational force and the (ii) resistive force of air is
$W_g = mgh = 10$ J; $W_a = \Delta K - W_g$.
$W_g + W_a = K_f - K_i$ $mgh + W_a = \frac{1}{2}mv^2 - 0$ $10^{-3}\times10\times10^{3} + W_a = \frac{1}{2}\times10^{-3}\times(50)^2$ $W_a = -8.75$ J, i.e. work done by air resistance is –8.75 J and work done by gravity is 10 J.
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