How Much Energy Was Lost to Friction When Braking Downhill?

Thread starter davedave
Start date Dec 5, 2008
Tags

Car Friction Work

AI Thread Summary

The discussion revolves around calculating the energy lost to friction when an 800kg car coasts down a 40m hill and slows from 6m/s to 20m/s. Participants are tasked with applying the conservation of energy principle, where the initial kinetic energy and potential energy must equal the final kinetic energy plus the energy lost to friction. The formula used is: initial kinetic energy + potential energy = final kinetic energy + friction energy. Different answers from participants indicate varying interpretations or calculations of the energy loss. The conversation highlights the importance of understanding energy conservation in physics problems involving friction and motion.

Dec 5, 2008

davedave

All my friends got different answers to the problem below. What is your answer?

An 800kg car moving at 6m/s begins to coast down a hill 40m high with its engine off. The driver applies the brakes so that the car's speed at the bottom of the hill is 20m/s. How much energy was lost to friction?

Physics news on Phys.org

Dec 5, 2008

mgb_phys

Science Advisor

Homework Helper

Kinetic energy_start + potential energy = kinetic energy_end + friction

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 »