11th Class Physics Unit 4 Work, Energy and Power Numericals with Solutions

4.1.A machine gun fires 6 bullets per minute with a velocity of 700 ms^{-1}. If each bullet has a mass of 40 g, then find power developed by the gun?

Given

Velocity V = 700 ms^-1

Mass per bullet m = 40g = 401000 = 0.04 kg

Firing rate and time t = 1 min = 60s

Formula

Kinetic energy of one bullet (K.E.) one bullet = 12mv²

Working

Calculate K.E. per bullet = 12 × 0.04 × (700)² = 9800 J

Total K.E. for 6 bullets (K.E.) six bullet = 6 × 9800 J

Formula

Power calculation P = wt = frac{(K.E.) six bullet{time

Result

Power developed = 6 × 980060 = 980 watts

4.2.A family uses 10 kW of power. Direct solar energy is incident on horizontal surface at an average rate of 300 W per square metre. If 75% of this energy can be converted into useful electrical energy, how large area is needed to supply 10 kW?

Given

Power required P_{required = 10 KW = 1000w

10 kW is 10,000 W, not 1,000 W. The book carries this slip through its working but its final answer, 44.4 m², is correct: 10,000 ÷ 225 = 44.4.

Solar energy incident rate I = PA = 300 watts m^-2

Efficiency Conversion efficiency = 75%

Formula

Useful power per unit area P_{useful = 75% × 300 watts m^-2 = 225 watts m^-2

Total area calculation Total area = frac{P_{required{P_{useful × area

Working

Area needed = frac{1000 w^-2}{225 watt m^-2 × 1m² = 44.4 m²

10 kW is 10,000 W, not 1,000 W. The book carries this slip through its working but its final answer, 44.4 m², is correct: 10,000 ÷ 225 = 44.4.

Total area = 44.4 m²

4.3.The mass of the Earth is 6.0 x 10^{24} kg and mass of the Sun is 1.99 x 10^{30} kg. The Sun is 160 million km away from the Earth. Find the value of gravitational P.E. of the Earth.

Given

Mass of Earth m = 6 × 10²⁴ kg

Mass of Sun m₂ = 1.99 × 10³⁰ kg

Distance r = 160 million km = 160 × 10⁶ × 10³ m = 1.6 × 10¹¹ m

Gravitational constant G = 6.674 × 10^-11 N m² kg^-2

Formula

Gravitational potential energy u = - Gm m₂r

Working

Substitute values = - frac{6.674 × 10^-11 × (6 × 10²⁴ × 1.99 × 10³⁰)}{1.6 × 10¹¹

Simplify numerator = - frac{6.674 × 10^-11 × 11.94 × 10⁵⁴{1.6 × 10¹¹

Continue calculation = - frac{79.69 × 10⁴³{1.6 × 10¹¹

Result

Gravitational P.E. = - 4.98 × 10³³ J

4.4.An object weighing 98 N is dropped from a height of 10 m. Its speed just before hitting the ground is 12 m s^{-1}. What is the frictional force acting on it?

Given

Weight W = 98 N

Mass m = Wg = 989.8 = 10kg

Height h = 10m

Speed just hitting ground v = 12 m s^-1

Formula

Energy balance equation Loss in P.E. = gain in K.E. + work done against friction

Energy equation mgh = 12mv² + fh

Working

Solve for friction work fh = 10 × 9.8 × 10 - 12 × 10 × 144

Calculate fh = 980 - 720 = 260 J

Result

Frictional force f = 26010 = 26N

4.5.A 75 watt fan is used for 8 hours daily for 30 days. Find: (i) Energy consumed in electrical units (ii) Electricity bill of fan of one unit costs Rs. 22.5?

Given

Power P = 75 W

Time per day t = 8 hours

Cost per unit cost/unit = Rs. 22.5

Formula

Energy consumption formula E = P(W) × t(h)1000 × days

Working

Calculate energy E = 75 × 8 × 301000 = 18 KWh

Formula

Electric bill calculation Electric bill cost = E × cost/unit

Result

Total cost = 18 × 22.5 = 405 Rs.

4.6.If an object of mass 2 kg thrown up from ground reaches a height of 5 m and falls back to the Earth (neglecting air resistance), calculate: (i) work done by gravity when the object reaches at 5 m height. (ii) work done by gravity when the object comes back to the Earth. (iii) Total work done by gravity in upward and downward motion. Also mention physical significance of the result.

Given

Mass m = 2kg

Height h = 5m

Formula

Work done = Force × Distance × cos(θ) w = Fd cos θ

(i) Work done by gravity during upward motion w_{rise = Fd cos 180^circ = (mg)(h) cos 180^circ = (2 × 9.8)(5)(-1) = -98 J

(ii) Work done by gravity during downward motion w_{fall = Fd cos 0^circ = (mg)(h) cos 0^circ = (2 × 9.8)(5)(1) = 98 J

(iii) Total work done by gravity w_{total = w_{rise + w_{fall = -98 + 98 = 0J

Physical significance Gravity is conservative, network over closed path is zero.

(i) w_{rise = -98 J, quad (ii) w_{fall = 98 J, quad (iii) w_{total = 0J

4.7.An electrical motor of one horse power is used to run a water pump. Water pump takes 15 minutes to fill a tank of 400 litres at a height of 10 m. Find (a) actual input work done by the electric motor to fill the tank. (b) actual output work done.

Given

Mass of water m = 400 litres = 400 kg (1 litre ≈ 1 kg)

Height h = 10m

Power P = 1hp = 746 watts

Time t = 15minutes = 900s

Formula

(a) Input work calculation w_{input = P · t = w_{input div t

Working

Calculate input work w_{input = pt = 746 × 900 = 671400 J

Formula

(b) Output work = Potential energy (PE) work output = m × g × h

Result

Output work = 400 × 9.8 × 10 = 39200 J

Note The work done by the motor is equal to the potential energy gained by the water, assuming no losses.

4.8.A passenger just arrived at the airport and dragging his suitcase to luggage check in desk. He pulls strap with a force of 200 N at an angle of 45° to the floor to displace it 50 m to the desk. Determine the value of work done by him on the suitcase.

Given

Force applied F = 200 N

Displacement d = 50 m

Angle θ = 45^circ

Formula

Work done formula Work done = Fd cos θ

Working

Calculate work = 200 × 50 × cos(45^circ)

Evaluate = 200 × 50 × 0.7071 = 7071 J

Work done = 7071 J

4.9.A1200 kg car is running at a speed of 40 km h^{-1}. How much power will be expended by it to accelerate at 2 ms^{-2}?

Given

Mass m = 1200kg

Velocity v = 40km/h = 40 × 10003600 = 11.11 ms^-1

Acceleration a = 2 m s^-2

Formula

Force calculation F = ma

Working

Calculate force = 1200 × 2 = 2400 N

Formula

Power formula Power = P = Fv

Result

Calculate power = 2400 × 11.11 = 26666.67 watts = 26.67k.w.

4.10.A200 g apple is lifted to 10 m and then dropped. What is its velocity when it hits the ground? Assume that 75% of work done in lifting the apple is transferred to K.E. by the time it hits the ground.

Given

Mass m = 200g = 0.2kg

Height h = 10m

Efficiency 75% of work transferred to K.E.

Formula

Work done by gravity w = Fd cos θ = (mg)(h) cos 0^circ = mgh cos 0^circ

Working

Calculate work = 0.2 × 9.8 × 10 × 1 = 19.6 J

K.E on hitting ground = 75(19.6) = 14.7 J

Formula

Kinetic energy formula K.E = 12mv²

Or equivalently v = sqrt{2(K.E)m

Working

Calculate velocity = sqrt{2(14.7)0.2 = sqrt{147} = 12.1 ms^-1

v = 12.1 m s^-1

Unit 4: Work, Energy and Power — Numericals