Question

How fast must a 6.0-kg cat run to have a kinetic energy of 150 J?

Answer 1

KE = 1/2 * mv^2
150 J = 1/2 * 6 kg * v^2
v^2 = (150 J) / (6 kg)(1/2)
= 50 m/s
V = sqrt(50 m/s)
5*sqrt(2)
~ 7.07 m/s

Answer 2

Answer: The cat runs at a speed of 7.07 m/s

Step-by-step explanation:

Kinetic energy is defined as the energy which is possessed due to its motion.

It is also defined as the half of the product of mass of the object and square of the velocity of the object.

Mathematically,

`E_K=(1)/(2)mv^2`

where,

`E_K` = kinetic energy of cat = 150 J

m = mass of the cat = 6.0 kg

v = velocity of cat

Putting values in above equation, we get:

`150J=(1)/(2)* 6.0* (v)^2\\\\v=\sqrt{(2* 150)/(6)}=7.07m/s`

Hence, the cat runs at a speed of 7.07 m/s