How far up the incline does the student go?
A student is sitting on a 10m high hill.A spring is compressed 50cm to launch a 100kg physics student. The track is frictionless until it starts up the incline. On the 30 degree incline the coeff of friction is 0.15. the spring constant is 80 000 N/m. the speed after the student launches from the spring is 14.1m/s. how far (in m) does the student go up the incline?
Answer:
Approach:
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Use conservation of energy: The kinetic energy KE is used up as the student goes up the incline; part of it goes into gaining gravitational potential energy PE and the rest represents work done against the friction Wf of the incline:
(1) KE = PE + Wf
Let d = the distance the student goes up the incline before he stops;
let h = the elevation at that point. Then
(2) sin(30) = h / d
First, the PE:
(3) PE = m * g * h
Second, the Wf, the work being done to overcome the frictional force Ff:
(4) Wf = Ff * d
The frictional force is given by:
(5) Ff = μk * Fn, where Fn is the force of gravity normal to the incline:
(6) Fn = m * g * cos(30)
Substituting (5) and (6) into (4):
(7) Wf = Ff * d
= (μk * Fn) * d
= (μk * m * g * cos(30)) * d
Substituting (7) and (3) into (1):
(8) KE = PE + Wf
= (m * g * h) + ((μk * m * g * cos(30)) * d)
= (m * g * d * sin(30)) + ((μk * m * g * cos(30)) * d) since from (2) h = d * sin(30)
= m * g * d * (sin(30) + μk * cos(30))
We know that the KE is given by:
(9) KE = 0.5 * m * v^2, since the track before the incline is frictionless (no losses).
Setting (9) equal to (8):
(10) 0.5 * m * v^2 = m * g * d * (sin(30) + μk * cos(30))
and solving for d, the distance up the ramp (along the ramp):
(11) d = (0.5 * m * v^2) / m * g * (sin(30) + μk * cos(30))
= (0.5 * v^2) / g * (sin(30) + μk * cos(30)) [cancelling the m’s]
Substituting values:
(12) d = (0.5 * 14.1^2) / 9.81 * (0.5 + 0.15 * 0.866)
= 99.4 / 6.18 = 16.1m up the incline <<===Answer
[The height (elevation) can be easily gotten from (2):
(13) h = d * sin(30) = 16.1 * 0.5 = 8.05m high]