27 Jun, 2021 06:36 PM

## 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?

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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]