Chapter 6, Problem 59P

An 85-kg football player traveling 5.0 m/s is stopped in 1.0 s by a tackier. (a) What is the original kinetic energy of the player? (b) What average power is required to stop him?

To determine

The original kinetic energy of the player.

Answer to Problem 59P

Solution:

Explanation of Solution

Given:

Mass of footballer $\text{=85 kg}$

Initial speed $\left(V_1\right)\text{=5}\text{.0m/s}$

Final speed $\left(V_2\right)\text{=0 m/s}$

Time, $t=1.0\text{s}$

Formula used: Kinetic energy $\left(K.E.\right)=\frac{1}{2}m{v}^2$

M : mass of body

V : speed of body

Work energy principle

  $\begin{array}{l}{W}_{net}=\Delta K=K_2-K_1\\ =\frac{1}{2}m{v}_2{}^2-\frac{1}{2}m{v}_1{}^2\end{array}$

Calculation:

  $\begin{array}{c}K.E=\frac{1}{2}m{v}_1{}^2\\ \text{ =}\frac{\text{1}}{\text{2}}\text{×85×}{\left(\text{5}\right)}^{\text{2}}\\ \text{ =1062}\text{.5 J}\\ \approx \text{1100 J}\end{array}$

Conclusion:The original kinetic energy of the player 1100 J.

To determine

What average power is required to stop him?

Answer to Problem 59P

Solution: 1100 WGiven: Mass of footballer $\text{=85 kg}$

Initial speed $\left(V_1\right)\text{=5}\text{.0}\text{m/s}$

Final speed $\left(V_2\right)\text{=0 m/s}$

Explanation of Solution

Formula Used:

  $\text{power=}\frac{\text{work}\left(\text{W}\right)}{\text{time}\left(\text{t}\right)}$

Calculation:

Work done by force required to stop him

  $\begin{array}{l}\text{=}\frac{1}{2}m{v}_1{}^2-\frac{1}{2}m{v}_2{}^2\\ \text{=}\frac{\text{1}}{\text{2}}\text{×}\left(\text{85kg}\right)\text{×}{\left(\text{0m/s}\right)}^{\text{2}}\text{-}\frac{\text{1}}{\text{2}}\text{×}\left(\text{85kg}\right)\text{×}{\left(\text{5}\text{.0m/s}\right)}^{\text{2}}\\ \approx \text{-1100 J}\end{array}$

Power $\text{=}\frac{\text{work}}{\text{time}}$

  $\begin{array}{l}\text{=}\frac{\text{1100}\text{J}}{\text{1}\text{.0s}}\\ \approx \text{1100 W}\end{array}$

Conclusion:Average power required to stop $85$ kg football player moving with moving $\text{95 km/h is 1100 W}$ .