Question

Certain force acting on 20 kg mass changes its velocity from 5 ms–1 to 2 ms–1.

Calculate the work done by the force.​

Answer 1

Answer:

Given, mass,

′

m=20kg;

initialvelocity,u=5ms

−1

;

finalvelocity,v=2ms

−1

Workdonebytheforce=changeinkineticenergy

or W=

2

1

​

mv

2

−

2

1

​

mu

2

=

2

1

​

m(v

2

−u

2

)

or W=

2

1

​

(20)[(2)

2

−(5)

2

]=−210J.

Explanation:

idk if this helped welp im a oof

Answer 2

Solution :-

As per the question ,

• Mass of the object = 20 kg

• Initial velocity = 5 m/s

• Final velocity = 2 m /s

We know that ,

⇒ W = Δ KE

⇒ W = KE f - KE i

here ,

• KEi = initial kinetic energy
• KEf = final kinetic energy

⇒ W = mv² / 2 - mu² / 2

⇒ W = m( v²-u²)/2

⇒ W = 20 ( 4 - 25 ) / 2

⇒ W = 10 x - 21

⇒ W = - 210 J

⇒ |W| = 210 J

The magnitude of work done by the object is 210 J