# Power/energy, calculate final velocity. Request for Assistance

#### Billyg

Having a brain cramp on this one. ...here is the question...a cart of a mass of 4.3kg moves along a horizontal track reaching a slope at 7.5m/s. As it rolls downhill to the bottom, 85J of heat energy is generated from friction until it reaches the bottom 3.6m below. Calculate the speed of the cart when it reaches the bottom of the track?
not sure if i am on the right track, as I am thinking that v is constant at 7.5m/s, height 3.6m, I calculated Kinetic energy horizontally is Ek=0.5mv^2.. or 0.5(4.3)(7.5^2)= 121J, then potential energy Ep=mgh or 4.3(9.8)(3.6)=152 J. so for total E= add all three Ek Ep Efr= so i am off track i think as i don't need Energy or even to use the P=W/T formula. I should be able to get away with v^2=Vin^2+2ad.. which is a= gravity 9.8m/s^2.. so is it just that simple to do V^2=7.5+2(9.8)(3.6)= V^2= sq root of 78.06 which is 8.83m/s is the final velocity as it reaches the bottom.. Total brain cramp... any help would be greatly appreciated.

#### Cervesa

initial kinetic energy + initial potential energy - work done by friction = final kinetic energy

$\dfrac{1}{2}mv_0^2 + mgh - W_f = \dfrac{1}{2}mv_f^2$

$\sqrt{v_0^2 + 2gh - \dfrac{2W_f}{m}}= v_f$

#### Billyg

Thanks, i think i got it now, i ended up getting 8.8m/s