Work And energy Numericals with Solutions

$p_1= m_1v_1$
$p_2= m_2 v_2$
But $p_1= p_2$ or
$m_1v_1 = m_2v_2$
If $m_1< m_2$ then $v_1> v_2$
$K_1 = \frac {1}{2}m_1v_1^2$
$K_2 = \frac {1}{2}m_1v_2^2$
From above equation
$K_1 = \frac {1}{2}p_1v_1$
$K_2 = \frac {1}{2}p_2v_1$
$K_1 : K_2 = v_1 : v_2$
But $v_1> v_2$
Therefore, $K_1 > K_2$

Potential Energy is given by
$PE= mgH$
Given mg=100N and H=15 m
$PE= 100 \times 15= 1500 J$

If If the same object is raised half of its original height i.e H= 7.5 m
$PE= 100 \times 7.5= 750 J$
If the same object is raised to three times of its original height i.e H=45 m
$PE= 100 \times 45= 4500 J$

1 Joule is the kinetic energy possessed by a body whose mass is 1 kg and moving with a velocity of $\sqrt {2}$ or 1.4 (approx) m/s.
$K = \frac {1}{2}mv^2$
For m=1 Kg and v=$\sqrt {2}$ m/s
$k=1 Joule$

Here m=2 Kg,u=10 m/s
$K = \frac {1}{2}mv^2$
Hence Kinetic energy at the time of throwing
$K = \frac {1}{2} \times 2 \times 10^2= 100 J$
When the Ball reaches at highest point, its whole kinetic energy is converted in Potential energy.
So potential energy of the ball at the highest point= Initial Kinetic energy=100J

$K_1 = \frac {1}{2}mv^2=200 J$
$K_2 = \frac {1}{2}(2m)v^2= 2 \times \frac {1}{2}mv^2=400 J$

m=500 Kg, v=36 km/hr=10 m/s
$K = \frac {1}{2}mv^2$
Hence
$K=25000 J= 25KJ$
When the velocity of the car is doubled
m=500 kg, v=20 m/s
Hence
$K=100000 J= 100 KJ$

Weight of the water =$800 \times 10 = 8000$ N
Height =1500 cm = 15 m
t=20 sec

$power = \frac {workdone }{time}$
$= \frac {mgh}{t} = \frac {8000 \times 15}{20} = 6000 Watt= 6 KW$

Power=60 W
Time=30 min= 1800 sec
$Energy= power \times time$
$= 60 \times 1800$
=108,000 J = 108 KJ

Power=1000 W = 1 KW
Time per day=2 hour
Total time= 2 X 28 = 56 hour
Energy= power X time
= 1 X 56
=56 KWH

Now 1 KWH costs Rs 3
Then 56 KWH will cost = 3 X 56 = Rs 168

a.Chemical energy into electric energy
b. electrical energy into mechanical energy

Two Bulb energy comsumed
$E_b= .04 \times 2 \times 6 = .48 KWH$
Two tubelights energy comsumed
$E_t= .05 \times 2 \times 8 = .8 KWH$
TV energy comsumed
$E_t= .12 \times 1 \times 6 = .72 KWH$
Total energy consumed per day
= .48 + .8 + .72 =2 KWH

Now Total bill will be given by
$=2 \times 30 \times 2.50 = Rs 150$

Weight of the car = 500 * 10 = 5000N
Height = 4 m
Workdone = $mgh= 5000 \times 4 = 20,000 J=20 KJ$

Power= Energy/time =500/20 = 25 Watt

Four Bulb energy comsumed
$E_b= .1 \times 4 \times 12 = 4.8 KWH$
5 fans energy comsumed
$E_f= .11 \times 5 \times 12 = 6.6 KWH$
Total Units consumed =4.8 + 6.6 = 11.4 KWH= 11.4 units
Now if one unit costs Rs 2.50
Total expenditure per day = 2.50 X 11.4 = Rs 28.5

The work done (W) in lifting a box through a height H against the gravitational force (F=mg) is given by W = FS = mgh

Hence , it is aparrant that it is independent of the rate at which the box is lifted

Geyser energy comsumed per day
$E_g= 2.5 \times 8 = 20 KWH$

Assuming month to be 31 days
Monthly consumption= 620 KWH =620 units
Cost of consumption = 3.50 X 620 = Rs 2170

Kinetic energy is given by
$K = \frac {1}{2}mv^2$
Now
$K_1=\frac {1}{2}mv^2$
$K_2=\frac {1}{2}m(3v)^2= 9 \times \frac {1}{2}mv^2$

$K_1 : K_2 =1: 9$

Kinetic Energy at the time of hitting the ground= Potential energy at the edge of cliff + Initial Kinetic energy
when the stond is throw vertically downward
$K_1 = mgh + \frac {1}{2}mv^2=1000J$ -(1)
Now when the stone is thrown horizontally outward with the same speed
$K_2 = mgh + \frac {1}{2}mv^2$
From equation
$K_2=1000 J$

Workdone to stop car = change in Kinetic energy = KE at stop - KE at start=0 -7000
=-7000 J (negative sign means work is done against car)

Power = Workdone /time = 7000/20 = 350 W

Zero. As displacement is zero.