A child and sled with a combined mass of 46.0 kg slide down a frictionless hill that is 10.4 m high at an angl?

if using kinematics: acceleration down the hill is g sin(theta) = 9.81m/s/s x sin 26 = 4.30m/s/s the distance down the hill is given by trigonometry = 10.4m/sin 26 = 23.7m therefore, using vf^2 = v0^2 + 2 a d we get vf^2 = 0+ 2 x 4.30m/s/s x 23.7m vf = 14.3m/s if using energy conservation, equate initial PE to final KE: m g h = 1/2 m v^2 v=Sqrt[2 g h] = Sqrt[2 x 9.81m/s/s x 10.4m] = 14.3m/s

draw the free body diagram of the scenario first..since no friction, no component of f=mu(R) in the opposing direction of motion..resolve the weight component. F=ma 46(9.81)sin26=46a a=46(9.81)sin26/46 you can now use suvat equations.. u=0 v= ?? a=46(9.81)sin26/46 s=10.4/tan26 v^2=u^2+2as v=sq.root(u^2+2as) substitute the values and get v .. hope you got that :D

I've been answering these questions all year at college.. I fail and cannot do this </3 **** MY LIFE.

draw the free body diagram of the scenario first..since no friction, no component of f=mu(R) in the opposing direction of motion..resolve the weight component. F=ma 46(9.81)sin26=46a a=46(9.81)sin26/46 you can now use suvat equations.. u=0 v= ?? a=46(9.81)sin26/46 s=10.4/tan26 v^2=u^2+2as v=sq.root(u^2+2as) substitute the values and get v .. hope you got that :D

I've been answering these questions all year at college.. I fail and cannot do this </3 **** MY LIFE.