Science And Technology Solutions Solutions for Class 9 Science Chapter 2 - Work And Energy

Answer:

a.

b. 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

or

Where, a is the acceleration of the body during the change in its velocity

Now, the work done on the body by the external force is given by:

W = F × s

F = ma …(ii)

From equations (i) and (ii), we obtain:

If the body was initially at rest (i.e., u = 0), then:

Since kinetic energy is equal to the work done on the body to change its velocity from 0 to v, we obtain:

c. 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
The object is stationary i.e. its initial velocity, u = 0.
Kinetic energy, $K=\frac{1}{2}m{v}^2=\frac{1}{2}m{u}^2=0$
Potential energy, U = mgh   .....(i)

At point B
Let the velocity of the object be v_B and the object has fallen through distance x. 
Using third equation of motion, we have
$$\begin{gathered} v_B^2=0+2gx \\ \Rightarrow v_B^2=2gx \end{gathered}$$
Kinetic energy, $K=\frac{1}{2}m{v}_B^2=mgx$
Potential energy, U = mg(h-x)    ......(ii)

At point C
Let the velocity of the object be v_C and the object has fallen through a distance h. 
Using third equation of motion, we have
$$\begin{gathered} v_C^2=0+2gh \\ \Rightarrow v_B^2=2gh \end{gathered}$$
Kinetic energy, $K=\frac{1}{2}m{v}_C^2=mgh$    ......(iii)
Potential energy, U = 0
From (i) and (iii), we see that the potential energy of the object at point A has transformed to its kinetic energy at point C. 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.

d. Work done, W = F$\times$S cos $\theta$
Here, $\theta$ = 30^o
Thus, W = F$\times$S cos 30 = $\frac{\sqrt{3}}{2}\times F\times S$

e. Momentum of an object, P = mv
If P = 0
$\Rightarrow v=0$
This is because mass of the object can never be 0. 
Now, kinetic energy of the object, K = $\frac{1}{2}\times m\times v^2$
Since, v = 0
$\Rightarrow K=0$
Hence, when the object has zero momentum, its kinetic energy is also zero.

f. Work done on an object is given as
W = F$\times$S cos $\theta$
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 $\theta$ between the force vector and displacement vector is always 90^o i.e.  $\theta$ = 90^o. Hence,
W = F$\times$S cos 90 = 0
Hence, the work done on an object moving in uniform circular motion zero.

Answer:

a. For work to be performed, energy must be transferred from one place to another.

b. Joule is the unit of work and energy.

c. 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 rapidity with which work is done

e. While dragging or lifting an object, the negative work is done by frictional and gravitational force, respectively.

Answer:

a. 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 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 will not change.

d. The work done on an object does not depend on initial velocity of the object.

Answer:

1. At the moment of releasing the balls, they posses potential energy.

2. The potential energy of the balls converts to kinetic energy as they roll down.

3. Note: The question is wrong. It should be "Why do the balls cover same distance after rolling down"?

The balls will posses same potential energy because the channels are of same height as well as the weight of the balls are same. So, the balls on reaching the bottom of their respective channels will posses same kinetic energy (following the law of conservation of energy). Thus, the balls will have same velocity at the bottom of the channel. Because of the same velocity, they will cover equal distance on the ground. Thus, both the balls covers equal distance after rolling down the incline because they possessed same potential energy at the top of the channels.

4. The total energy of the ball will eventually be in the form of kinetic energy.

5. Law of conservation of energy is demonstrated using this activity. At the top of the channels, the total mechanical energy of the balls is in the form of potential energy. Now, as the balls roll down, they come into motion. This motion shows that the balls now posses kinetic energy too. As law of conservation of energy states that energy cannot be created nor destroyed, this means this kinetic energy came into existence because of decrease in potential energy of the ball (decrease in potential energy is due to decrease in height). Also, when the ball reaches the bottom, the total mechanical energy of the balls is in the form of kinetic energy. This shows that all the potential energy of the balls got converted to kinetic energy at the bottom.  

Answer:

a. Power, P = 2 kW
$$\begin{gathered} \mathrm{Power}, P=\frac{\mathrm{Work} \mathrm{done} \left(W\right)}{\mathrm{Time} \mathrm{taken} \left(T\right)} \\ \Rightarrow W=P\times T \end{gathered}$$
For every minute i.e. 60 s, work done by the pump is
$W=2\times 1000\times 60=120000 \mathrm{J}$
Now, this work done is stored in water as its potential energy. Thus,
$$\begin{gathered} mgh=120000 \mathrm{J} \\ \Rightarrow m=\frac{120000}{9.8\times 10}=1224.5 \mathrm{kg} \end{gathered}$$
Hence, 1224.5 kg of water is lifted by the pump every minute to a height of 10 m.

b. Total electricity consumed in the month of April is
$$\begin{gathered} E=P\times T\times (\mathrm{Total} \mathrm{days} \mathrm{in} \mathrm{April} \mathrm{month}) \\ E=1200\times 30\times 60\times 30=64.8\times {10}^6 \mathrm{J} \\ \mathrm{We} \mathrm{know}, \\ 1 \mathrm{unit} =3.6\times {10}^6 \mathrm{J} \\ \mathrm{Thus}, \\ E=18 \mathrm{units} \end{gathered}$$

c. Energy of the ball at the height of 10 m = 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 = $\frac{60}{100}\times 10mg=6mg$
or, mgh' = 6mg =  6 m

d. Here, v = 72 km/h = 20 m/s, u = 54 km/h = 15 m/s, m = 1500 kg
Work done by the car = Change in kinetic energy of the car
i.e. 
$$\begin{gathered} W=\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}-\mathrm{K}.{\mathrm{E}}_{\mathrm{i}} \\ W=\frac{1}{2}\times m\times v^2-\frac{1}{2}\times m\times u^2=\frac{1}{2}\times m(v^2-u^2) \\ =\frac{1}{2}\times 1500({20}^2-{15}^2)=131250 \mathrm{J} \end{gathered}$$

e. Work done, W = F$\times$S$\times$cos $\theta$
Here, $\theta$ = 0^o
F = 10 N
S = 30 cm = 0.03 m
$\therefore W=10\times .03\times 1=3 \mathrm{J}$