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?

AB
27 Jun, 2021 06:36 PM

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]

LE
27 Jun, 2021 06:37 PM

kinetic + grav. potential coming down the ramp = grav. potential going up ramp + friction force

1/2*m*v^2+m*g*h=D*m*g*sin(theta)+D*m*g*u*cos(theta)