Question

the velocity of a 2 kg body is changed from (4i+3i)ms^-1 to 6m/s the work done on the body is​

Answer 1

Answer:

hope it will help you

Explanation:

Given : m

1

=2kg m

2

=1kg u

1

=3 m/s u

2

=4 m/s

Let the common velocity of the combined body be V

Mass of combined body M=2+1=3kg

Applying conservation of momentum :

2×3−1×4=3×V ⟹V=

3

2

m/s

Momentum of the system P=2×3−1×4=2 kg m/s

Loss in kinetic energy of the system due to collision ΔK.E=K.E

i

−K.E

f

ΔK.E=

2

1

(2)×3

2

+

2

1

(1)×4

2

−

2

1

(3)×(

3

2

)

2

⟹ΔK.E=

3

49

J

Answer 2

Concept:

• Work done on a body
• Acceleration and force acting on a body
• Vectors
• Calculating the magnitude of a vector
• Kinetic energy

Given:

• The initial velocity of the body u = 4i + 3j m/s
• The final velocity of the body v = 6 m/s
• The mass of the body m = 2 kg

Find:

• The work done on the body

Solution:

The magnitude of the initial velocity of the body u = √4²+3² = √25 = 5 m/s

The speed increased.

So the kinetic energy also increased. This increase can only be due to work done on the body.

Therefore, the difference in the initial and the final kinetic energies is equal to the work done on the body.

Initial KE = 1/2mu²

Final KE = 1/2mv²

Work done on the body = Final KE - Initial KE

Work done on the body = 1/2mv² - 1/2mu²

Work done on the body = 1/2m (v² - u²)

Work done on the body = 1/2 (2) (6² - 5²)

Work done on the body =  6² - 5²

Work done on the body =  36- 25

Work done on the body =  11 J

The work done on the body is 11 J.

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