Q. 5 D

# Solve the following examples.

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 ?

Answer :

Now here velocity of a car is increased i.e. car is accelerated which means some force is applied on the car and since car is in motion and accelerating so it also travelled some displacement from original position

Situation has been depicted in the figure

Now we know force on a body is given as

F = ma

Where F force is applied on a body of mass m, to move it with acceleration a

we know work done on an object due to a force is given by relation

W = FS

Where W is the work done due to force F in displacing object by S in the direction of force

So putting the value of force in above equation we get the magnitude of work done as

W = FS = maS

Mass of car is given

m = 1500 kg

Now, finding the value of (aS)

We know newton’s third equation of motion which is

v^{2} – u^{2} = 2aS

Here v is final velocity of object, u is the initial velocity of the object, S is the displacement covered by object and a is its acceleration during motion

here velocities are given in Km/hr , converting them to SI i.e. m/s

we know , 1 Km = 1000 m

1 hour=60 minute,1 minute = 60 sec

So, 1 hour = 60 × 60 = 3600 sec

So we get,

1 km/hr = 1000 m/3600 s

=10/36 = 5/18 m/s

Now here initial velocity of the car is

u = 54 km/hr = 54 × 5/18 = 15 m/s

final velocity of the Car is

v = 72 km/hr = 72 × 5/18 = 20 m/s

putting values of v and u in above equation we get

(20ms^{-1})^{2} - (15ms^{-1})^{2} = 2×aS

i.e.

so putting the value of mass of car and value of aS in equation of work done we get work done as

W = maS = 1500kg × 87.5 m^{2}s^{-2} = 131250 J

So the amount of work done is 131250 J

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Rewrite the following sentences using proper alternative.

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.

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