Question

The speed of a moving body is doubled. What is the change in its kenetic energy

Answer 1

K = mv^2 /2

K propotional v^2

when v becomes 2v

k' = 4k

kinetic energy rises 4 times

Answer 2

Answer:

4 times

Explanation:

According to the given question, we are asked to find the kinetic energy of a body if speed is doubled.

So,

• Let v = initial speed

• v' = final speed = 2v

(According to question)

• m = mass of the body

• k = initial kinetic energy

• k' = final kinetic energy

We know that,

Kinetic energy = 1/2 mv^2

• So, k = 1/2 mv^2-----------(1)

• And, k' = 1/2 mv'^2----------(2)

Dividing equation (1) and (2) we get,

(As the mass is constant and 1/2 is just a numerical value then it will be cancelled out when being divided)

(And thus we will get this equation as shown below)

• K'/K = v'^2/ v^2

• K'/K = 4v^2/v^2

(As v' = 2v)

• K'/K = 4

• K' = 4K

Hence,

Final kinetic energy = 4× Initial Kinetic energy

So, kinetic energy will be 4 times of initial kinetic energy if velocity of the moving body is being doubled from the start.