Work and Energy
Based on Maharashtra Board Class 9-Science-Chapter-2-Notes-Solution-Video-Test and PDF
Solution
Question 1:
Write detailed answers?
1. Explain the difference between potential energy and kinetic energy.
Potential Energy
Kinetic Energy
It is possessed by a body by virtue of its configuration and position.
It is possessed by a body by virtue of its motion.
It occurs in various forms such as gravitational potential energy and electric potential energy but work is not done till it is transformed into kinetic energy.
It occurs only in one form and for doing work it does not have to be transformed into another form.
Gravitational potential energy = mgh.
Kinetic energy =1/2 mv2
otential energy can be negative
Kinetic energy cannot be negative.
eg: A book kept on a shelf has potential energy.
eg: A moving car posses kinetic energy.
2. Derive the formula for the kinetic energy of an object of mass m, moving with velocity v.
Let the initial velocity of the object be u. Let an external force be applied on it so that it gets displaced by distance s and its velocity becomes v. In this scenario, the kinetic energy of the moving body is equal to the work that was required to change its velocity from u to v. Thus, we have the velocity−position relation as: v2 = u2 + 2as v2 - u2 = 2as \(s=\frac{v^2-u^2}{2a}\) ………… (i) Where, a is the acceleration of the body during the change in its velocity If F is the force acting on the body, the work done on the body by the external force is given by: W = F × s ……….. (ii) F = ma …(iii) From equations (i) (ii) and (ii), we obtain: W=ma x \(\frac{v^2-u^2}{2a}\) = \(\frac{1}{2}m(v^2-u^2)\) If the body was initially at rest (i.e., u = 0), then: W = \(\frac{1}{2}mv^2\) Since kinetic energy is equal to the work done on the body to change its velocity from 0 to v, we obtain: Kinetic Energy of the body = \(\frac{1}{2}mv^2\)
3. Prove that the kinetic energy of a freely falling object on reaching the ground is nothing but the transformation of its initial potential energy.
Let the object be at point A i.e. at height h above the surface of Earth as shown in the figure below.
At point A At point B ⇒ v12=2gx At point C ⇒ = v12 =2gh \ Total energy =mgh + 0 = mgh ......(iii)
The object is stationary i.e. its initial velocity, u = 0.
Kinetic energy, KE= \(\frac{1}{2}mu^2\) = 0
Potential energy, PE = mgh
Total Energy = 0 + mgh =mgh .....(i)
Let the velocity of the object be v1 and the object has fallen through distance x.
Using third equation of motion, we have
v12=u2+2gx = 0+2gx …….( u=0)
Kinetic energy, KE= \(\frac{1}{2}m(v_1)^2\) = \(\frac{1}{2}m2gx\) =mgx
Potential energy, PE = mg(h-x) ......(ii)
Total energy = mgx + mgh – mgx = mgh .....(ii)
Let the velocity of the object be v2 and the object has fallen through a distance h.
Using third equation of motion, we have
v22=0+2gh …… (u=0)
Kinetic energy, KE = \(\frac{1}{2}\)mv22 =mgh
Potential energy, PE = 0 at ground level
From (i) (ii) and (iii), Total energy at point A, B and C are same i.e total energy remains constant. Thus, it can be concluded that the kinetic energy of a freely falling object on reaching the ground is nothing but the transformation of its initial potential energy.
4. Determine the amount of work done when an object is displaced at an angle of 30o with respect to the direction of the applied force.
Work done, W = F×s cosθ cos30o = \(\frac{\sqrt{3}}{2}\) Thus, W = F×s cos30 = \(\frac{\sqrt{3}}{2}\)Fs
Here, θ = 30o
5. If an object has 0 momentum, does it have kinetic energy? Explain your answer.
Momentum of an object, P = mv
If P = 0
⇒v=0
This is because mass of the object can never be 0.
Now, kinetic energy of the object, K = 1/2×m×v2
Since, v = 0
⇒K=0
Hence, when the object has zero momentum, its kinetic energy is also zero.
6. Why is the work done on an object moving with uniform circular motion zero?
Work done on an object is given as
W = F×s cos θ
In circular motion, the direction of force acting on the object is radially inward and the direction of motion of the object is tangential to the circular path at every instant of time. Thus, the angle θ between the force vector and displacement vector is always 90o i.e. θ = 90o. and cos90 = 0 Hence,
W = Fs cos90 = 0
Hence, the work done on an object moving in uniform circular motion zero.
Question 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
For work to be performed, energy must be transferred from one place to another.
b. Joule is the unit of ...
(i) force
(ii) work
(iii) power
(iv) energy
Joule is the unit of work and 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
The gravitational force and the reaction force in vertical direction have same magnitude. Friction is not action as the horizontal surface is smooth.
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
Power is a measure of the rapidity with which work is done
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
While dragging or lifting an object, the negative work is done by frictional and gravitational force, respectively.
Question 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
(ii) sitting on the ground
(iii) sleeping on the ground
(iv) standing on the ground
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
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.
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.
The work done on an object does not depend on initial velocity of the object.
Question 4:
Study the following activity and answer the questions.
- Take two aluminium channels of different lengths.
- Place the lower ends of the channels on the floor and hold their upper ends at the same height.
- 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?
Potential energy.
2. As the balls roll down which energy is converted into which other form of energy?
The potential energy of the balls converts to kinetic energy as they roll down.
3. Why do the balls cover the same distance on rolling down?
They have same speed
4. What is the form of the eventual total energy of the balls?
The total energy of the ball will eventually be in the form of kinetic energy.
5. Which law related to energy does the above activity demonstrate? Explain.
Law of conservation of energy is demonstrated using this activity.
Question 5:
Solve the following examples.
1. An electric pump has 2 kW power. How much water will the pump lift every minute to a height of 10 m?
Power, P = 2 kW =2 x 1000 w= 2000 w Power, P = Work done (W)/Time taken (t) ⇒ W = P×t For every minute i.e. t= 60 s, work done by the pump is W=2000w×60s=120000 J Now, this work done is stored in water as its potential energy. Thus mgh=120000 J ⇒m=120000/9.8×10=1224.5 kg Hence, 1224.5 kg of water is lifted by the pump every minute to a height of 10 m.
2. If a 1200 W electric iron is used daily for 30 minutes, how much total electricity is consumed in the month of April?
t=30daysx30min/day x 60s/min=54000s Total electricity consumed in the month of April is W=1200×54000=64.8×106 J We know,1 unit = 1kW-h= 3.6×106 J ∴ W= 64.8×106 / 3.6×106 Thus, W =18 units Electricity consumed = 18 units
W=P×t
3. If the energy of a ball falling from a height of 10 metres is reduced by 40%, how high will it rebound?
h= 10mtr At height h 10 mtr, Energy of the ball = mgh = 10mg Let the ball rebounds to a height h' where the energy reduces by 40%. Thus, Energy at height h' = mgh' = 60% of energy at height of 10 m ∴ mgh’ = (60/100)×10mg=6mg ∴ mgh' = 6mg ∴ h’ = 6m The ball will rebound to to a height of 6m.
4. The velocity of a car increase from 54 km/hr to 72 km/hr. How much is the work done if the mass of the car is 1500 kg?
Here, v = 72 km/h = 20 m/s, u = 54 km/h = 15 m/s, m = 1500 kg W=(1/2)×m×v2−(1/2)×m×u2 =(1/2)×m(v2−u2)=(1/2)×1500(202−152) =131250 J Work done by car = 131250 J
Work done by the car = Change in kinetic energy of the car
i.e.
W=K.Ef − K.Ei
5. 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?
Work done, W = F×s×cos θ Work done by Ravi= 3 J
Here, θ = 0o
F = 10 N
s = 30 cm = 0.3 m
∴ W = 10 × 0.3 × 1 = 3 J
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