# A physics student performs an experiment to measure the efficiency of a go-cart. From rest, he rolls a cart of total mass 95 kg from the top of a 3.8 m high hill. After rolling to the bottom of the hill, his go-cart reaches a speed of 6.7 m/s. a) What is the total energy of the student and go-cart before he begins the run? b) What is the kinetic energy of the student and go-cart at the bottom of the ramp? c) What is the efficiency of the system?

(A) While the cart is at the top of the ramp but before it starts moving, it has zero kinetic energy K = 0 and its total energy is potential energy, V. (Why is potential energy represented as V? I honestly have no idea, but this is the standard notation. I've also seen U, but for some reason, never P.)

That potential energy is entirely due to gravity, so it is given by `V = m g h`. This system dissipates energy into its environment (hence why its efficiency is not simply 100%), so we should really distinguish initial energy `E_i ` from final energy `E_f`.

`E_i = K_i + V_i`

`E_i = 0 + mgh = (95 kg)(9.80 m/s^2)(3.8 m)`

`E_i = 361 J`

We should give our answer in two significant figures since that's what we were given, so `E_i = 360 J`.

(B) At the bottom of the ramp, the cart has lost all its potential energy, but now has kinetic energy given by `K = 1/2 mv^2`. This is its new total energy.

`E_f = K_f + V_f`

`E_f = 1/2mv^2 + 0 = 1/2(95 kg)(6.7 m/s)^2 `

`E_f = 318.25 J`

Once again, round to two significant figures: `E_f = 320 J`

(C) Efficiency in the sense we're using it here is just final energy divided by total energy:` e = E_f / E_i = (320 J)/(360 J) = 0.89` This system has an efficiency of 89%. That makes sense; we would expect it to be a bit less than 100%.

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