Given below are the Class 9 Science CBSE Numericals for Work And energy
a. Concepts questions
b. Calculation problems
c. Multiple choice questions
d. Long answer questions
e. Fill in the blank's
Question 1.A light and a heavy object have the same momentum. Find out the ratio of their kinetic energies. Which one has a larger kinetic energy?
Answer
$p_1= m_1v_1$
$p_2= m_2 v_2$
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, $K_1 > K_2$
Question 2.An object with 100 N weights is raised to a height of 15m. Find the potential energy possessed by the object at that height. Also find the new potential energy:
a.If the same object is raised half of its original height.
b.If the same object is raised to three times of its original height?
Given g=10m/s
^{2}
Answer
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
$PE= 100 \times 45= 4500 J$
Question 3.Define one joule of kinetic energy?
Answer
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
$k=1 Joule$
Question 4.A ball of mass 2kg is thrown up with a speed of 10m/s. find the kinetic energy of the ball at the time of throwing. Also find the potential energy of the ball at the highest point?
Answer
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
Question 5.An object of mass m and velocity v has kinetic energy= 200J. Find the new kinetic energy if the mass of the object becomes double and velocity still remains the same?
Answer
$K_1 = \frac {1}{2}mv^2=200 J$
$K_2 = \frac {1}{2}(2m)v^2= 2 \times \frac {1}{2}mv^2=400 J$
Question 6.Calculate the kinetic energy of a car of mass 500kg moving with a velocity of 36km/h. Find the kinetic energy if the velocity of car doubles?
Answer
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
$K=100000 J= 100 KJ$
Question 7.Calculate the power of an electric motor that can lift 800 kg of water to store in a tank at a height of 1500cm in 20s. (g=10m/s
^{2}).
Answer
Weight of the water =$800 \times 10 = 8000$ N
Height =1500 cm = 15 m
t=20 sec
$power = \frac {workdone }{time}$
$= \frac {mgh}{t} = \frac {8000 \times 15}{20} = 6000 Watt= 6 KW$
Question 8.Calculate the electrical energy consumed in Joules if a toaster of 60 W is used for 30 minutes?
Answer
Power=60 W
Time=30 min= 1800 sec
$Energy= power \times time$
$= 60 \times 1800$
=108,000 J = 108 KJ
Question 9.An electric heater of 1000 W is used for two hours in a day? What is the cost of using it for a month of 28 days, if one unit costs 3.00 rupees?
Answer
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
Question 10.What energy transformation takes place in :
a.Dry cell
b.electric fan
Answer
a.Chemical energy into electric energy
b. electrical energy into mechanical energy
Question 11.Calculate the electricity bill amount for a month of 30 days, if the following devices are used as specified:
a.two bulbs of 40W for six hours.
b.two tubelights of 50W for eight hours
c.A TV of 120W for six hours.
d.Give the rate of electricity is 2.50 rupees per unit?
Answer
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
$=2 \times 30 \times 2.50 = Rs 150$
Question 12.A crane pulls up a car weighing 500kg to a vertical height of 4m. Calculate the work done by the crane?
Given g=10m/s
^{2}
Answer
Weight of the car = 500 * 10 = 5000N
Height = 4 m
Workdone = $mgh= 5000 \times 4 = 20,000 J=20 KJ$
Question 13.A lamp consumes 500 J of electrical energy in 20 sec. What is the power of the lamp?
Answer
Power= Energy/time =500/20 = 25 Watt
Question 14.In a factory four bulbs of 100 W each and five fans of 110 W each operate for 12 hours daily. Calculate the units of electricity consumed? Also find the total expenditure if one unit costs 2.50 rupees?
Answer
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
Question 15.The work done in lifting a box on to a platform does not depend upon how fast it is lifted up. Explain your answer giving proper reason?
Answer
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
Question 16.A geyser of 2.5kW is used for eight hours daily? Calculate the monthly consumption of electrical energy units. Also calculate the cost of electrical units consumption in a month if rate per unit is 3.50 rupees?
Answer
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
Question 17.Two bodies of equal masses move with the uniform velocities v and 3v respectively. Find the ratio of their kinetic energy?
Answer
Kinetic energy is given by
$K = \frac {1}{2}mv^2$
Now
$K_1=\frac {1}{2}mv^2$
$K_2=\frac {1}{2}m(3v)^2= 9 \times \frac {1}{2}mv^2$
$K_1 : K_2 =1: 9$
Question 18. A boy stands on the edge of a cliff and throws a stone vertically downward with an initial speed of 10 m/s. The instant before the stone hits the ground below, it has 1000 J of kinetic energy. If he were to throw the stone horizontally outward from the cliff with the same initial speed of 10 m/s, how much kinetic energy would it have just before it hits the ground?
Answer
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
$K_2=1000 J$
Question 19.
The kinetic energy of a car is 7000 J as it travels along a horizontal road. How much work is required to stop the car in 20 s?
Answer
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
Question 20. A 1.5 m high person is holding a 20 kg trunk on his head and is standing at a roadways bus-terminal. How much work is done by the person?
Answer
Zero. As displacement is zero.
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