Question

A 60 kg runner in a sprint moves at 11 m/s. A 60 kg cheetah in a sprint moves at 33 m/s. By what factor does the kinetic energy of the cheetah exceed that of the human runner?

Answer 1

Explanation:

It is given that,

Mass of the runner, m₁ = 60 kg

Speed of runner, v₁ = 11 m/s

Mass of cheetah, m₂ = 60 kg

Speed of cheetah, v₂ = 33 m/s

Kinetic energy of runner, $E_r=\dfrac{1}{2}m_1v_1^2$

$E_r=\dfrac{1}{2}\times 60\times (11)^2=3630$........(1)

Kinetic energy of cheetah, $E_c=\dfrac{1}{2}m_2v_2^2$

$E_c=\dfrac{1}{2}\times 60\times (33)^2=32670\ J$..............(2)

From equation (1) and (2), it is clear that

$E_c= 9\times E_r$

So, the kinetic energy of the cheetah is nine times that of human runner. Hence, this is the required solution.

Answer 2

Answer:

The kinetic energy of the cheetah exceed by the 9 times of the runner.

Explanation:

Given that,

Mass of runner = 60 kg

Velocity of runner = 11 m/s

Mass of cheetah = 60 kg

Velocity of cheetah = 33 m/s

We need to calculate the kinetic energy of the cheetah

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Kinetic energy of runner

$K.E_{r}=\dfrac{1}{2}m_{r}v_{r}^2$....(I)

Kinetic energy of cheetah

$K.E_{c}=\dfrac{1}{2}m_{c}v_{c}^2$....(II)

From equation (I) and (II)

$\dfrac{K.E_{r}}{K.E_{c}}=\dfrac{\dfrac{1}{2}m_{r}v_{r}^2}{\dfrac{1}{2}m_{c}v_{c}^2}$

$K.E_{c}=\dfrac{\dfrac{1}{2}m_{c}v_{c}^2}{\dfrac{1}{2}m_{r}v_{r}^2}\timesK.E_{r}$

Put the value into the formula

$K.E_{c}=\dfrac{\dfrac{1}{2}\times60\times(33)^2}{\dfrac{1}{2}\times60\times(11)^2}\timesK.E_{r}$

$K.E_{c}=9K.E_{r}$

Hence, The kinetic energy of the cheetah exceed by the 9 times of the runner.