Distance of student up the incline?

The spring is compressed 50 cm and used to launch a 100 kg physics student downhill. The track is frictionless until it starts up the incline. The student's coefficient of kinetic friction on the 30° incline is 0.17. The student's speed just after losing contact with the spring is 14.282 m/s.





(b) How far up the incline does the student go?


wrong check mark m (measured along the incline)

Distance of student up the incline?
The students energy when leaving the spring is:


E = KE + PE


E = 1/2 m v^2 + mgh


E = 0.5*100*14.282^2 + 100*9.8*h


E = 10198.7762 + 98h





At the bottom of the hill he will have a velocity of:


KE = 1/2 m v^2


10198.7762 + 98h = 0.5*100*v^2


v = sqrt(204 + 1.96 h)





the frictional force acting against him will be:


f = uN


f = 0.17*100*9.8*cos(30)


f = 14.4





causing a deceleration of


F = ma


14.4 = 100a


a = 0.144 m/s^2





so the student will travel up the ramp





a = delta_v/delta_t


-0.144 = (0-sqrt(204 + 1.96 h))/detla_t


0.0208*delta_t^2 = 204+196h


delta_t^2 = 9800 + 9415.5h





d = V_i*delta_t + 1/2*a*delta_t^2


d = sqrt(204 + 1.96 h)*sqrt(9800 + 9415.5h) + 0.5*0.144*(9800 + 9415.5h)





where h is the hight of the hill at which the student was released