Numerical Problems

Oct 12, 2022

A man has pulled a cart through 35 m applying a force of 300 N. Find the work done by the man. (10500 J)

Difficulty: Easy

Solution:       Distance = S = 35 m                      Force = F = 300 N                      Work done = W = ?                                    W = $F \times S$                                    W = $300\times 35$                                    W = 10500 J

A block weighing 20 N is lifted 6 m vertically upward. Calculate the potential energy stored in it. (120 J)

Difficulty: Easy

Solution:        Weight of the block = W = 20 N                       Height = h = 6 m                       Potential energy = P. E. =?                                 P.E. = mgh We know that           w = mg                                 P.E. = $(mg) \times h$Thus,                         P.E. = $(2 \times 10)\times 6$                                  P.E. = 120 J

A car weighing 12 k N has a speed of 20 $ms^{-1}$. Find its kinetic energy. (240 kJ)

Difficulty: Medium

Solution:       Weight of the car = w = 12kN = $12\times 1000$ N = 12000 NSpeed of the car = v= 20 $ms^{-1}$ Kinetic energy = K.E. = ? K.E. = $\frac{1}{2} m v ^{2}$W = mg or m = $\frac{w}{g}$m = $\frac{12000}{10}$ = 1200 kg Thus, K.E. = $\frac{1}{2}\times 1200 \times (20)^{2}$= $600 \times 400$ = 240000 JK.E. = 240 kJ

A 500 g stone is thrown up with a velocity of $15 ms^{-1}$. Find its (i) P.E. at its maximum height (ii) K.E. when it hits the ground(56.25 J, 56.25 J)

Difficulty: Medium

Solution:     Mass of stone = m = 500 g = $\frac{500}{1000}$ kg = 0.5 kg Velocity = v = 15 $ms^{-1}$  Potential energy = P.E. = ? Kinetic energy = K.E. = ? Loss of K.E. = Gain in P.E.   $\frac{1}{2}m v_\_{f}^{2}$ – $\frac{1}{2}m v\_{i}^{2}$  = mgh As the velocity of the stone at maximum height become zero, therefore, $v_{f}$  = 0 $\frac{1}{2}\times 0.5\times (0) -\frac{1}{2}\times 0.5 \times (15)^{2}$ = mgh $\frac{-1}{2}\times 0.5 \times 225$ = mgh $-$56.25 = mgh mgh = $-$56.25J   Since energy is always positive, therefore P.E. = 56.25 J K.E. = $\frac{1}{2}m v ^{2}$ K.E. = $\frac{1}{2}\times 0.5 \times (15)^{2}$ = $\frac{1}{2}\times 0.5 \times 225$ = 56.25 J

On reaching the top of a slope 6 m high from its bottom, a cyclist has a speed of 1.5 $ms^{-1}$. Find the kinetic energy and the potential energy of the cyclist. The mass of the cyclist and his bicycle is 40 kg. (45 J, 2400 J)

Difficulty: Easy

Solution:   Height of the slope = h = 6 m                  Speed of the cyclist = v = 1.5 $ms^{-1}$                   Mass of cyclist and the bicycle = m = 40 kg            Kinetic energy = K.E. = ?            Potential energy = P.E. = ?            (i) K.E. = $\frac{1}{2}m v ^{2}$    K.E = $\frac{1}{2}\times 40\times (1.5)^{2}$    K.E = $\frac{1}{2}\times 40\times 2.25$ = 45 J    (ii) P.E = mgh     P.E = $40 \times 10 \times 6$ = 2400 J

A motor boat moves at a steady speed of 4 $ms^{-1}$. Water-resistance acting on it is 4000 N. Calculate the power of its engine. (16 kW)

Difficulty: Easy

Solution:               Speed of the boat = v = $4ms^{-1}$                          Force = F = 4000 N                         Power = P = ?                        P = F v                       P = $4000 \times 4$                       P = 16000 W                       P = $16 \times 10^{3}$ W                       P = 16 kW

A man pulls a block with a force of 300 N through 50 m in 60 s. Find the power used by him to pull the block. (250 W)

Difficulty: Easy

Solution:                                 Force = F = 300 N                        Distance = S = 50m                        Time = t = 60 s                        Power = P = ?                          Power =$\frac{work}{time}$ = $\frac{w}{t}$= $\frac{(F\times S)}{t}$                         P =$\frac{(300\times50)}{60}$ = $5 \times50$ = 250 W

A 50 kg man moved 25 steps up in 20 seconds. Find his power, if each step is 16 cm high. (100 W)

Difficulty: Easy

Solution:          Mass = m = 50 kgTotal height = h = $25 \times 16$ = 400 cm = $\frac{400}{100}$ m = 4 mTime = t = 20 sPower = P = ? Power =$\frac{work}{time}$=$\frac{w}{t}$ = $\frac{mgh}{t}$P =$\frac{(50\times 10\times 4)}{20}$P = 100 W

Calculate the power of a pump that can lift 200 kg of water through a height of 6 m in 10 seconds. (1200 watts)

Difficulty: Easy

Solution:          Mass = m = 200 kg                     Height = h = 6m                     Power = P = ?                      Power =$\frac{work}{time}$ = $\frac{w}{t}$ = $\frac{mgh}{t}$                     P = $\frac{(200 \times 10\times 6)}{10}$                      P = 1200 W

An electric motor of 1hp is used to run the water pump. The water pump takes 10 minutes to fill an overhead tank. The tank has a capacity of 800 liters and a height of 15 m. Find the actual work done by the electric motor to fill the tank. Also, find the efficiency of the system. (Density of water = 1000 $kgm^{-3}$ )(Mass of 1 liter of water = 1 kg) (447600 J, 26.8%)

Difficulty: Hard

Solution:             Power = P = 1 hp = 746 W                   Time = t = 10 min = $10\times 60$ s = 600 s                   $\frac{Capacity}{volume}$ = V = 800 liters                   Height = h = 15 m   Work done = W = ? Efficiency = E = ?   Power = $\frac{work}{time}$              P =$\frac{w}{t}$   Or          W = $P \times t$               W = $746\times 600$               W = 447600 J   Since the work done by the electric pump to fill the tank is 447600 J. It is equal to the input. Hence input = actual work doe = W = 447600 J   Output = P.E = mgh Since                 1 litre = 1kg, therefore 800 litres = 800 kg Output = P.E = $800 \times 10\times 15$ = 120000J                         % Efficiency    = $\frac{Output}{Input }×100$                   % Efficiency  =  $\frac{120000}{447600}×100$                       Efficiency = 26. 8 %

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