### Numerical Problems and Solutions of Power

Formulas and Solutions to Numerical Problems of Power

Important Formula of Power:

1.) $P=\frac{Work(W)}{Time(t)}$

2.) $P=\frac{F.s}{t}$

3.) $P=\frac{F \: s \: cos \theta}{t}$

4.) $P=\frac{F \: s }{t} \qquad \left( \because \theta =0^{\circ} \right)$

5.) $P=F \: v \qquad \left( \because v = \frac{s}{t} \right)$

Numerical Problems and Solutions

Q.1 A woman pulls a bucket of water of mass $5 Kg$ from a well which is $10 m$ deep in 10 sec. Calculate the power used by her $(g=10 \: m/sec^{2})$.

Solution:
Given that:
The mass of bucket of water $(m)=5 \: Kg$ Depth of well $(h)=10 \: m$

The time taken to pull a bucket from well $t= 10 \: sec$

The value of gravitational acceleration $g=10 \: m/sec^{2}$

The power used $(P)=?$

Now the power used by her:

$P=\frac{W}{T}$

$P=\frac{mgh}{t} \qquad \left( \because W=mgh \right)$

Now Substitute the given values in the equation of power:

$P=\frac{5 \times 10 \times 10}{10}$

$P=50 \: Joule/sec=50 W$

Q.2 A man whose mass is $50 \: Kg$ climbs up $30$ steps of the stairs in $30 \: Sec$. If each step is $20 \: cm$ high, Calculate the power used in climbing the stairs $(g=10 \: m/sec^{2})$

Solution:
Given that:
Mass of a man $(m) = 50 \: Kg$

Climbs up the number of Steps $(N) = 30$

The time is taken to climb up the 30 steps $(t)= 30 \: sec$

The length or height of one step $(l)=20 cm \: = \: 0.20 \: m$

The total length or height of 30 steps $h=30 \times 0.20 \: m$

The power used in climbing the stairs=?

Now from the equation of power:

$P=\frac{W}{t}$

$P=\frac{mgh}{t} \qquad \left( \because W=mgh \right)$

Now Substitute the given values in the above equation:

$P=\frac{50 \times 10 \times 30 \times 0.20 }{30}$

$P=100 \: Joule/sec = 100 \: W$

Q.3 A horse exerts a pull of $300 N$ on a cart so that the horse-cart system moves with a uniform speed of $18 Km/h$ on a level road. Calculate the power in watt developed by the horse and also find its equivalent in horsepower.

Solution:
Given that:
The horse exerts the pull i.e force $(F)=300 N$

The unifrm speed of the horse-cart $(v)=18 Km/h$

So the distance moved in $(s)=18 Km=18000m$

The time $(t)=1h= 60 \times 60 sec$

The power developed by the horse$(P)=?$

From the equation of the power, the power developed by horse in one hour:

$P=\frac{W}{t}$ $P=\frac{F.s}{t}$

Now Substitute the given values in the above equation:

$P=\frac{3000 \times 18000}{60 \times 60}$

$P=1500 W$

$P=\frac{1500}{746}$

$P=2 \: hp$

Q.4 A man weighing $60Kg$ climbs up a staircase and carrying a load of $20 Kg$ on his head. The staircase has 20 steps each of height $0.2m$. If he takes $10 sec$ to climb, find his power.

Solution:
Given that:
The weight of man $(m_{1})=60Kg$

The weight of load $(m_{2})=20Kg$

The number of steps in staircase $(N)=20$

The height of each step $(H)=10s$

The total mass of the man and load $(m)=m_{1}+m_{2}= 60+20=80 Kg$

The total heght of stair case $(h)=N \times H = 20 \times .2= 4 m$

The power of man: $P=\frac{W}{t}$

$P=\frac{mgh}{t}$

$P=\frac{80 \times 9.8 \times 4}{10}$

$P=313.6 W$

Q.5 A car of mass $2000 Kg$ and it is lifted up a distance of $30m$ by a crane in $1 \: min$. A second crane does the same job as first crane in $2 \: min$. Do the both cranes consume the same or different amounts of fuel? Find the power supplied by each crane? Neglecting power dissipation against friction.

Solution:
Given that:
The mass of car $(m)=2000 Kg$

The lifted up distance $(h)=30m$

The time taken by first crane $(t_{1}= 1\: min)$

The time taken by second crane $(t_{2}= 2\: min)$

The work done by each crane:

$W=mgh$

$W=2000 \times 9.8 \times 30$

$W=5.88 \times 10^{5} J$

As both the cranes do the same amount of work, both consume the same amount of fuel.

The power supplied by the first crane:

$P_{1}=\frac{mgh}{t_{1}}$

$P_{1}=\frac{2000 \times 9.8 \times 30}{60}$

$P_{1}=9800 W$

The power supplied by the second crane:

$P_{2}=\frac{2000 \times 9.8 \times 30}{1.2}$

$P_{2}=4900 W$

Q.6 The human heart discharges $75 \: mL$ of blood at every beat against a pressure of $0.1 m$ of Hg. Calculate the power of the heart assuming that pulse frequency is $80$ beats per minute. Density of $Hg=13.6 \times 10^{3} Kg/m^{3}$.

Solution:
Given that:
The volume of blood discharge per beat $(V)=75 \: mL = 75 \times 10^{-6} m^{-3}$

The pressure of blood $(P)=0.1 m \: of \: Hg$ i.e.

$P= 0.1 \times 13.6 \times 10^{3} N-m^{-2} \quad (\because P=\rho g h)$

The work done per beat $=PV$

The workdone in 80 beats$W =80 \times PV$

The power of heart:

$P=\frac{80 \times PV}{60}$

$P=\frac{80 \times 0.1 \times 13.6 \times 10^{3} \times 75 \times 10^{-6} \times 9.8}{60}$

$P=1.33 W$

Q.7 A machine gun fires $60$ bullets per minute with a velocity of $700 m/sec$. If the mass of each bullet is $50 g$ then find the power developed by the gun.

Solution:
Given that:
The mass of one bullet $M=50 g$

The number of bullet $N=60$

The mass of $60$ bullets $m=M \times N= 50\times 60 = 300g= 3Kg$

The velocity of the bullet $v=700 m/sec$

The time take to fire $60$ bullets $t=1 \: min = 60 sec$

The power developed by gun:

$P=\frac{W}{t}$

According to the work-energy theorem:

$W=\Delta K$

$P=\frac{\Delta K}{t}$

$P=\frac{m v^{2}}{2t} \quad( \because K=\frac{mv^{2}}{2})$

$P=\frac{3 \times (700)^{2}}{2 \times 60}$

$P=12250 W$