9 Class Work and Energy exercise question answer science and technlogoy
Class 9 science and technology chapter 2 question answer work and energy | Work and energy class 9 exercise answers Maharashtra board | Work and energy class 9 solution state board.
1. Write detailed answers?
a. Explain the difference between potential energy and kinetic energy.
Ans:
|
Potential Energy |
Kinetic Energy |
|
1.
The energy possessed by a body on account of its specific state or position
is called its potential energy. |
1. The energy
possessed by a body due to its motion is called its kinetic energy. |
|
2.It occurs
various forms such as gravitational potential energy and electric potential
energy but work is not done till it is transformed into kinetic energy. |
2.It occurs only in one form and for
doing work it does not have to be transformed into another form. |
|
3.Gravitational
potential energy = mgh. |
3. Kinetic energy = ½
mv2 |
|
4.
Potential energy can be negative. |
4. Kinetic energy cannot be negative. |
b. Derive the formula for the kinetic energy of an object of mass m, moving with velocity v.
Ans:
Consider
a body of mass m moving with a uniform acceleration a along a straight line. If u is the initial
velocity of the body, V is the final velocity and s is the distance covered by
the body during this change of velocity we have
V2 = u2 + 2 as ( the kinematical equation of motion)
V2 – u2 = 2 as
S = V2 – u2 / 2a ………..(1)
If F is the net force acting on the body and W is the work done by the force,
W = Fs ……….(2)
Now, F = ma ………..(3)
From Eqs. (1), (2) and (3) we get,
W = ma × V2 – u2 / 2a
= 1/2 m (V2 – u2 )
= For a body initi ally at rest, u = 0
W = 1/2
Work done on a body = change in the kinetic energy of the body.
Kinetic
energy of the body. = 1/2 mv2
c. Prove that the kinetic energy of a freely falling object on reaching the ground is nothing but the transformation of its initial potential energy.
Ans:
Consider a body of mass m initially at rest at point A at a hight h from the ground . suppose that the body is released so that if follows the downward path ABC freely ( i.e., the buoyancy due to air and air resistance are ignored).
- At A
u (initial velocity) = 0
Kinetic energy (KE) = 1/2 mu2 = 0
Potential energy (PE) = mgh, where g is the acceleration due to gravity.
Total energy = 0 + mgh = mgh
- At B
- At C
If v is the velocity of the body at C, v2 = u2 + 2gh
(considering the motion from A to C)
v2 = 2 gh (u = 0)
KE = ½ mv2
= ½ m (2 gh) = mgh
PE = 0 at the ground level
Therefore
the total energy of the body is the same
at points A, B and C, i.e., the total energy remains constant during the motion
of the body.
d. Determine the amount of work done when an object is displaced at an angle of 300 with respect to the direction of the applied force.
Ans:
e. If an object has 0 momentum, does it have kinetic energy? Explain your answer. f. Why is the work done on an object moving with uniform circular motion zero?
Ans:
Kinetic energy of a body of mass m and speed v
= K = ½ mv2 = m2v2 / 2m
= P2/2m, where p = mv is the momentum of the body,
If
follows that if p =0 , the kinetic energy of the body is zero
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2. Choose one or more correct alternatives.
a. For
work to be performed, energy must be ….
(i) transferred from one place to another
(ii) concentrated
(iii) transformed from one type to another
(iv) destroyed
Ans:
(i) transferred from one place to another
(iii)
transformed from one type to another
b.
Joule is the unit of …
(i) force
(ii) work
(iii) power
(iv) energy
Ans:
(i) force
(iv)
energy
c.
Which of the forces involved in dragging a heavy object on a smooth, horizontal
surface, have the same magnitude?
(i) the horizontal applied force
(ii) gravitational force
(iii) reaction force in vertical direction
(iv) force of friction
Ans:
(ii) gravitational force
(iii)
reaction force in vertical direction
d.
Power is a measure of the …….
(i) the rapidity with which work is done
(ii) amount of energy required to perform the work
(iii) The slowness with which work is performed
(iv) length of time
Ans:
(i) the rapidity with which work is done
(iii)
The slowness with which work is performed
e.
While dragging or lifting an object, negative work is done by
(i) the applied force
(ii) gravitational force
(iii) frictional force
(iv) reaction force
Ans:
(ii) gravitational force
(iii)
frictional force
3. Rewrite the following sentences using proper alternative.
a. The potential energy of your body is least when you are …..
(i) sitting on a chair
(iii) sleeping on the ground
(ii) sitting on the ground
(iv) standing on the ground
Ans: The potential energy of your body is least when you are sleeping on the ground.
b. The total energy of an object falling freely towards the ground …
(i) decreases
(ii) remains unchanged
(iii) increases
(iv) increases in the beginning and then decreases
Ans: The total energy of an object falling freely towards the ground remains unchanged
c. If
we increase the velocity of a car moving on a flat surface to four times its
original speed, its potential energy ….
(i) will be twice its original energy
(ii) will not change
(iii) will be 4 times its original energy
(iv) will be 16 times its original energy.
Ans: If we increase the velocity of a car moving on a flat surface to four times its original speed, its potential energy will not change.
d. The work done on an object does not depend on ….
(i) displacement
(ii) applied force
(iii) initial velocity of the object
(iv) the angle between force and displacement.
Ans: The work done on an object does not depend on initial velocity of the object.
4. Study the following activity and answer the questions.
1. Take two aluminium channels of different lengths.
2. Place the lower ends of the channels on the floor and hold their upper ends at the same height.
3. Now take two balls of the same size and weight and release them from the top end of the channels. They will roll down and cover the same distance
Questions
1. At
the moment of releasing the balls, which energy do the balls have?
Ans:
Potential energy
2. As
the balls roll down which energy is converted into which other form of energy?
Ans:
potential energy – Kinetic energy
3. Why
do the balls cover the same distance on rolling down?
Ans: They have the same speed
4. What
is the form of the eventual total energy of the balls?
Ans:Kinetic energy
5.
Which law related to energy does the above activity demonstrate ? Explain.
Ans: Law of conservation of energy .
Energy
can neither be created nor destroyed. IT can be converted from one form into
another. Thus, the total amount of energy in the universe always remains
constat.
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5. Solve the following examples.
a. An electric pump has 2 kW power. How much water will the pump lift every minute to a height of 10 m? (Ans : 1224.5 kg)
Ans:
Given Data: P = 2 k W = 2000 J/s, t= 60 s,
h = 10 m , g = 9.8 m/s2, m=?
P = W/t = mgh/t
m = pt/gh = 2000 J/s × 60 s ÷ 9.8 m/s × 10 m
= 12000 / 9.8 kg = nearly 1224 kg
This
gives the required mass of water.
b. If a 1200 W electric iron is used daily for 30 minutes, how much total electricity is consumed in the month of April? (Ans :18 Unit)
Ans:
Given Data : P = 1200 W,
t = 30 days × (30 minutes / day ) × ( 60 s / minute ) = 54000 s,
W = ?
P = W / t
W = pt
W = 1200 W × 54000 s
= 648 × 105 J
= 6.48 × 107 J
= 6.48 × 107 J ÷ 3.6 × 106 Units
= 18 units
This
gives the required electricity.
c. If the energy of a ball falling from a height of 10 metres is reduced by 40%, how high will it rebound? (Ans : 6 m)
Ans:
Given Data: h1 = 10 m.
E2 = E1 – E2 = E1 – E1 × 40/100
E2 = E1 ( 1 – 0.4 ) = 0.6 E1, h2 = ?
E1 = mg h1 , E2= mg h2 ,
E2 = 0.6 E1
h2 = 0.6 E1 = 0.6 × 10 m = 6 m
The
ball will rebound to a height of 6 m .
d. The velocity of a car increases from 54 km/hr to 72 km/hr. How much is the work done if the mass of the car is 1500 kg ? (Ans. : 131250 J)
Ans:
Given Data: m = 1500 kg , u = 54 km/h
= 5400m/3600s = 15 m/s,
V = 72 km/h = 72000m/3600s
= 20 m/s, W = ?
Work done,
W = increase in kinetic energy
= ½ mv2 – ½ mu2
= ½ m (v2 – u2)
= ½ × 1500 kg × [ (20 m/s)2 – (15 m/s)2]
= 750 × (400 – 225)
J = 750 × 175 J
= 131250 J
The
work done to increase the velocity of the car = 131250 J
e. Ravi applied a force of 10 N and moved a book 30 cm in the direction of the force. How much was the work done by Ravi? (Ans: 3 J)
Ans:
Given Data: F = 10 N, s = 30 cm = 0.3 m, W = ?
W = Fs = 10 N × 0.3 m = 3 J
The work done by Ravi = 3 J
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