What is the original kinetic energy of the player?

Thread starter needhelp83
Start date Oct 28, 2006
Tags

Energy Kinetic Kinetic energy

AI Thread Summary

The original kinetic energy of a 95-kg football player traveling at 5.0 m/s is calculated to be 118,750 Joules using the formula K=(1/2)mv². To stop the player in 1.0 second, the average power required is also determined to be 118,750 Watts, applying the power formula P=w/t. There is a discussion on the relevance of converting kilograms to grams, with some questioning the necessity of this conversion. The calculations align with the work-energy principle, confirming their accuracy. Overall, the discussion emphasizes the relationship between kinetic energy, power, and the work-energy principle in physics.

Oct 28, 2006

needhelp83

A 95-kg football player traveling 5.0 m/s is stopped in 1.0 s by a tackler.
a. What is the original kinetic energy of the player?
b. What average power is required to stop him?

a.
K=(1/2)mv2
K=(1/2)(9500 g)(5.0 m/s)2= 118,750 J
b.
P=w/t
P=118750 J/1.0 s= 118750 W

Would this be correct because of the work-energy principle?

Physics news on Phys.org

Oct 28, 2006

radou

Homework Helper

Looks okay, but I don't see why you should convert [kg] to [g].

Thread 'I've come across another fluid pressure problem I don't understand'

Neglect gravity other than that of water; neglect friction. F1=ρgsh=1000*10*1*(1+4)=50000 N; F1=ρgsh=1000*10*0.1*4=4000 N; F3=50000-4000=46000 N?

View full post »

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

View full post »

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

View full post »