Work, Energy and Power — ICSE Class 10 Physics Important Questions

Answer: (c) zero.

The force (weight) is vertical while the displacement is horizontal, so $\theta=90^\circ$θ=90° and $W=Fs\cos90^\circ=0$W=Fs cos 90°=0. No work is done against gravity.

Answer: (b) $3.6\times10^{6}\text{ J}$3.6×10⁶ J.

$1\text{ kWh}=1000\text{ W}\times3600\text{ s}=3.6\times10^{6}\text{ J}$1 kWh=1000 W×3600 s=3.6×10⁶ J.

Answer: (c) four times.

$KE=\dfrac12 mv^2$KE=1/2 mv², so $KE\propto v^2$KE∝ v². Doubling $v$v multiplies $KE$KE by $2^2=4$2²=4.

Answer: (d) horsepower.

Horsepower is a unit of power: $1\text{ hp}=746\text{ W}$1 hp=746 W. The others are all units of energy: $1\text{ J}=10^{7}\text{ erg}$1 J=10⁷ erg, $1\text{ eV}=1.6\times10^{-19}\text{ J}$1 eV=1.6×10^-19 J and $1\text{ kWh}=3.6\times10^{6}\text{ J}$1 kWh=3.6×10⁶ J.

Answer: (a) In a circular orbit the gravitational (centripetal) force points to the centre while the velocity is tangential, so $\theta=90^\circ$θ=90° and $W=Fs\cos90^\circ=0$W=Fs cos 90°=0. R correctly explains A.

Work is done when a force acts on a body and moves it in the direction of the force; it equals the product of the force and the displacement in the direction of the force, $W=F\,s\cos\theta$W=F s cos θ. Its SI unit is the joule (J).

Power is the rate of doing work, $P=\dfrac{W}{t}$P=W/t. Its SI unit is the watt (W), where $1\text{ W}=1\text{ J s}^{-1}$1 W=1 J s^-1.

Work done $=$= force $\times$× height:
$W=F\times h=750\text{ N}\times16\text{ m}=12000\text{ J}.$W=F× h=750 N×16 m=12000 J.

Power:
$P=\frac{W}{t}=\frac{12000\text{ J}}{5\text{ s}}=2400\text{ W}=2.4\text{ kW}.$P=W/t=(12000 J)/(5 s)=2400 W=2.4 kW.

(a) The pull makes an angle $\theta=60^\circ$θ=60° with the displacement:
$W=F\,s\cos\theta=50\times10\times\cos60^\circ=50\times10\times\tfrac12=250\text{ J}.$W=F s cos θ=50×10× cos 60°=50×10× 1/2=250 J.

(b) Zero. The weight acts vertically downward while the displacement is horizontal, so $\theta=90^\circ$θ=90° and $W=mg\,s\cos90^\circ=0$W=mg s cos 90°=0.

(c) Friction acts opposite to the displacement, so $\theta=180^\circ$θ=180°:
$W=20\times10\times\cos180^\circ=-200\text{ J}.$W=20×10× cos 180°=-200 J.

The work done by friction is negative.

Kinetic energy:
$KE=\frac12 mv^2=\frac12\times5\times(10)^2=\frac12\times5\times100=250\text{ J}.$KE=1/2 mv²=1/2×5×(10)²=1/2×5×100=250 J.

By the work-energy theorem, the work done by the retarding force equals the loss in kinetic energy:
$F\times s=250\text{ J}$F× s=250 J
$F\times25=250\Rightarrow F=10\text{ N}.$F×25=250⇒ F=10 N.

The retarding force is $10\text{ N}$10 N.

Principle of conservation of energy: Energy can neither be created nor destroyed; it only changes from one form to another, so the total energy of an isolated system remains constant.

Consider a body of mass $m$m dropped from height $H$H (take $g$g constant, no air resistance).

At the top ($h=H$h=H, $v=0$v=0): $PE=mgH$PE=mgH, $KE=0$KE=0, total $=mgH$=mgH.

At a point after falling a distance $x$x (height $H-x$H-x): velocity $v^2=2gx$v²=2gx.
$KE=\tfrac12 mv^2=\tfrac12 m(2gx)=mgx,\qquad PE=mg(H-x).$KE=1/2 mv²=1/2 m(2gx)=mgx, PE=mg(H-x).
$\text{Total}=mgx+mg(H-x)=mgH.$Total=mgx+mg(H-x)=mgH.

At the ground ($h=0$h=0): $v^2=2gH$v²=2gH, so $KE=mgH$KE=mgH, $PE=0$PE=0, total $=mgH$=mgH.

At every point the total mechanical energy equals $mgH$mgH, hence it is conserved.

(i) Energy in kWh:
$E=P\times t=2\text{ kW}\times2.5\text{ h}=5\text{ kWh}\ (=5\text{ units}).$E=P× t=2 kW×2.5 h=5 kWh (=5 units).

(ii) Energy in joules:
$E=5\text{ kWh}\times3.6\times10^{6}\text{ J kWh}^{-1}=1.8\times10^{7}\text{ J}.$E=5 kWh×3.6×10⁶ J kWh^-1=1.8×10⁷ J.

Cost $=$= units $\times$× rate $=5\times\text{Rs }6=\text{Rs }30.$=5× Rs 6=Rs 30.

(a) Work-energy theorem: The net work done by all the forces acting on a body equals the change in its kinetic energy, $W=\Delta KE=\tfrac12 mv^2-\tfrac12 mu^2$W=Δ KE=1/2 mv²-1/2 mu².

(b) Kinetic energy is the energy possessed by a body due to its motion, e.g. a moving car. Potential energy is the energy possessed by a body due to its position or configuration, e.g. water stored in a dam at a height.

(c) Initial kinetic energy:
$KE=\tfrac12 mv^2=\tfrac12\times0.5\times(20)^2=\tfrac12\times0.5\times400=100\text{ J}.$KE=1/2 mv²=1/2×0.5×(20)²=1/2×0.5×400=100 J.

By conservation of energy, at the highest point all KE becomes PE:
$PE_{\text{max}}=100\text{ J}.$PE_max=100 J.

Maximum height:
$mgh=100\Rightarrow h=\frac{100}{0.5\times10}=20\text{ m}.$mgh=100⇒ h=100/0.5×10=20 m.

Given $m=3000\text{ kg}$m=3000 kg, $h=20\text{ m}$h=20 m, $v=4\text{ m s}^{-1}$v=4 m s^-1, $t=10\text{ min}=600\text{ s}$t=10 min=600 s.

(a) Potential energy gained:
$PE=mgh=3000\times10\times20=6\times10^{5}\text{ J}=600000\text{ J}.$PE=mgh=3000×10×20=6×10⁵ J=600000 J.

(b) Kinetic energy given:
$KE=\tfrac12 mv^2=\tfrac12\times3000\times(4)^2=\tfrac12\times3000\times16=24000\text{ J}.$KE=1/2 mv²=1/2×3000×(4)²=1/2×3000×16=24000 J.

(c) Total energy supplied $=PE+KE=600000+24000=624000\text{ J}$=PE+KE=600000+24000=624000 J.
$P=\frac{\text{total energy}}{t}=\frac{624000}{600}=1040\text{ W}=1.04\text{ kW}.$P=(total energy)/t=624000/600=1040 W=1.04 kW.

The power of the pump is $1040\text{ W}$1040 W.

(i) At the extreme position the bob is momentarily at rest, so its energy is entirely gravitational potential energy.

(ii) Potential energy at the top:
$PE=mgh=0.2\times10\times0.45=0.9\text{ J}.$PE=mgh=0.2×10×0.45=0.9 J.

(iii) At the lowest point all PE converts to kinetic energy:
$\tfrac12 mv^2=0.9\text{ J}\Rightarrow v^2=\frac{2\times0.9}{0.2}=9\Rightarrow v=3\text{ m s}^{-1}.$1/2 mv²=0.9 J⇒ v²=2×0.9/0.2=9⇒ v=3 m s^-1.

(iv) In the absence of friction the total mechanical energy ($KE+PE$KE+PE) stays constant at $0.9\text{ J}$0.9 J throughout the swing; only the form of energy changes between PE and KE.