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Question

A sled carrying a monkey is given an initial speed of 5.0 m/s up a 20 degree inclined plane. the combined mass of the sled+monkey is 20 kg. The coefficient of kinetic friction betwwen the sled and the inclined plane is 0.20. How far up the incline do the monkey and sled move?

Answer 1

Initial velocity of sled is (Vi) 5 m/s .

Inclination of the plane is given by 20°

The mass of sled and monkey is m 20 kg .

The coefficient of kinetic friction (μk)  0.20 .

The  normal force is F_N =  mg cos ø 

The force with  kinetic friction is 

F_F = μk F_N

F_F = μk mg cosø 

F_F = (0.20) (20)(9.8) cos 20

F_F = 36.835 N .

So the force with kinetic friction is 36.835 N .

The net force accelerating the sled up the ramp, F_a, is the difference between the component of the sled’s weight along the ramp and the frictional force opposing it:

F_a  =   mg sin ø - F_F 

F_a  =  (20)(9.8) sin 20 - 36.835

F_a  =  67.035 -  36.835

F_a  =  30.2 N 

So the net force is 30.2 N .

Acceleration of  sled (a) = F_a /m                                 [ since F = ma ]

a = 30.2 / 20 

a = 1.51 m/s² .

The final velocity is (Vf)  =  0                                          

The distance travel by  sled can be evaluted can be found using  

S = [(Vf)² - (Vi)²]/2 a                                                [ Since (Vf)² - (Vi)² = 2aS   ]

S = [ 0² - 5² ] / 2(1.51)

S = -25 / 2(1.51)

S = -8.278 

Here the negative sign indicates that sled is moving up  .

So the distance travel by  sled  is 8.278 m