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**SHORT ANSWER TYPE QUESTIONS**

**Q.1. A truck of mass 1800 kg is moving with a speed 54 km/h. When brakes are applied, it stops with uniform negative acceleration at a distance of 200 m. Calculate the force applied by the brakes of the truck and the work done before stopping. [CBSE 2016]****Ans:** Given: Mass, m = 1800 kg; Velocity,

u = 54 km/h = m/s = 15 m/s; v = 0;

Distance, 5 = 200 m

Retardation, a

=

Force, F = ma = 1800 x

= -1012.5 N

The negative sign indicates force acts in the opposite direction to motion.

Work done = Fs = 1012.5 x 200

= 202500 J**Q.2. Define power. [ CBSE 2016]****Ans: **The rate of doing work or the rate of transfer of energy is known as the power.

∴ **Q.3. State the law of conservation of energy. [CBSE 2016]Ans:** Energy can neither be created nor be destroyed. It can only be transformed from one form to another.

(b) The masses of scooters and bikes are in the ratio of 2 :3, but both are moving with the same speed of 108 km/h. Compute the ratio of their kinetic energy. [CBSE 2016]

Ans:

(b) Kinetic energy ∝ Mass of body

Let mass of scooter = m

and mass of bike = m

∴

∴ Total force applied by labourer F = (m

Work done W = Fs = (m

When the labourer carries an additional load of mass m

F' = (m

Total work done W' - F's = (m

= (50 + 60 + 10) x 10 x 15 = 18000 J

Ans:

= 10 m s

= Work to be done on the car - Increase in K.E. of car

⇒

After a time t = 2 s, the body is at a height h, where

∴ Potential energy of body at t = 2 s, E

(а) how would the potential energy of the car change?

(b) When velocity of car is doubled v'/ = 2v) then its kinetic energy E

(c) Momentum 'p' of car is given as : p = mv. When velocity of the car is doubled, its momentum is also accordingly doubled to p' = 2p

p' = mv' - m(2v) = 2mv - 2p

(Take g = 10 m/s

(b) A driver speeds up his vehicle when he moves up a hill. Give reason. [CBSE 2016]

∴ Energy supplied to turbine = Loss in potential energy of water = mgh

∴

(b) When a driver moves his vehicle up a hill, the force of gravity opposes his motion and he needs greater power. For this purpose, the driver speeds up his vehicle.

∴ Power of the lamp P =

Further electrical energy consumed by the lamp in operation for t = 2 hours daily for n = 15 days is

Ans:

Or

State the relation between kW h and joule. [CBSE 2015]

Ans:

(i) ∴ Work was done on the body against the force of gravity.

W = Fs = mg x h - 0.1 x 10 x 10 - 10 J

(ii) The potential energy of the body E

Ans:

Commercial unit of energy: kilowatt-hour (kWh)

SI unit of energy: joule (J)

Given: power, P = 400 W = 0.4 kW;

Time, t = 2 h

Energy unit consumed in a day = Pt

= 0.4 kW x 2 h = 0.8 kWh

Ans:

(i) its mass, and (ii) its height above the ground level.

When a person lifts an object up, two forces are acting on the object.

These are :

(i) Force due to gravity F - mg, where m = mass of the object and g = acceleration due to gravity. The force acts in vertically downward direction.

(ii) A force F applied by the person whose magnitude is equal to F but direction is opposite. Thus, F' = F = mg acts in vertically upward direction.

Out of these two forces (a), the force F applied by the person does positive work because the displacement of object in upward direction is along the direction of force F. (b) The force of gravity F = mg does negative work because the force F - mg acts in vertically downward direction and displacement is upward.

(a) K.E. of obiect on reaching ground E

(b) ∵ , hence velocity v =

Ans:

E

and energy consumed in the factory 5 fans of 70 W each for 12 hours

E

∴ Total electricity consumed E = E

∴ Total expenditure @ Rs. 2.00 per unit = 2 x 10.2 = Rs. 20.4

(b) A person carrying a load of 20 kg climbs 4 m in 10 seconds. Calculate the work done and his power, (g = 10 m/s

Ans:

= 2 m/s

(b) Given: Mass, m = 20 kg; Height, h = 4 m; Time, t = 10 s; g = 10 ms

Work done, W = mgh

= 20 x 10 x 4 = 800 J

and

**LONG ANSWER TYPE QUESTIONS**

**Q.1. (a) Define work. Give SI unit of work. Write an expression for positive work done.****(b) Calculate the work done in pushing acai through a distance of 50 m against the force of friction equal to 250 N. Also state the type of work done.****(c) Sarita lives on 3rd floor of building at the height of 15 m. She carries her school ba weighting 5.2 kg from the ground floor to her house. Find the amount of work done by her and identity the force against which she has done work (g = 10 ms ^{-2}) [CBSE 2016]**

Its SI unit is joule (J).

Positive work done, W = F.s,

where F = force; 5 = displacement is direction of force.

(b) Given: Distance, s = 50 m; Force, f = 250

N (opposite to direction of friction)

Work, W = F .s = 250N x 50m = 12500 J Work done by applied force is positive and by friction is negative.

(c) Given: Mass, m = 5.2 kg; g = 10 m s

Work done = mgh = 5.2 x 10 x 15 = 780 J

Work is done against the force of gravity acting on the bag.

Consider an object of mass m in a state of motion with an initial velocity u. Let now a constant force F acts on it and displaces the body through a distance s in the direction of force applied.

∴ Work was done on the object W = Fs

Due to the work done on the body, let velocity of the object changes from u to v and a be the acceleration produced.

Then according to the equation of motion

v

⇒

Again according to second law of motion,

we have F = ma

∴ Work was done on the object W = Fs = (ma)

If the object started from rest, then u = 0 and hence W =

The work done on the object is equal to the kinetic energy imparted to the object. Thus, the kinetic energy possessed by an object of mass m moving with a uniform velocity v is given by

(b) Here mass of ball m = 400 g - 0.4 kg and speed of ball v = 25 m s

∴ Kinetic energy of ball E

(a) for using 3 bulbs of 40 W each of 6 hours

(b) for using 4 tube lights of 50 W each for 6 hours

(c) for running a refrigerator of 320 W for 24 hours

∴ Total energy consumed per day E - E

∴ Total energy consumed during a month of 30 days = 30 x 9.60 kW h = 288 units

∴ Electricity bill @ Rs. 2.50 per unit - Rs. 2.50 x 288 = Rs. 720

∴ Work (W) = Constant force applied (F) x Displacement along the direction of force (s).

⇒ SI unit of work is called joule (JorMjm). Work is said to be 1 joule if under the influence of a force of 1 N the object moves} through a distance of 1 m along the direction of applied force.

(b) Here mass of truck m = 3000 kg, initial speed u = 72 km/h

final speed u = 0 and force applied by brakes F - 24000 N

As force of brakes opposes motion we take it negatively, that is F = 24000 N

∴ Acceleration of truck a =

Using the relation v

Distance covered

and work done by the force W = Fs = (- 24000) x 25 = - 600000 J = -6x 10

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