kinetic friction and acceleration physics question

Question

A student of mass 57.4 kg, starting at rest, slides down a slide 23.2 m long, tilted at an angle of 20.1° with respect to the horizontal. If the coefficient of kinetic friction between the student and the slide is 0.113, find the force of kinetic friction, the acceleration, and the speed she is traveling when she reaches the bottom of the slide. (Enter the magnitudes.)

 

 

Answer 1

Hi Jessica

Net downward force along the plane            ( force of kinetic friction fk= μMg Cosθ )

F= MgSinθ - μMg Cosθ  

 So acceleration 

 =( gSinθ - μg Cos θ)  

Where θ =20.1 , μ =0.113

M= 57.4 kg

 

V=  √2( gSinθ - μ gCosθ ) S  Where S = 23.2 m  V = velocity at the bottom of the plane

 

g = 9.8 m/s2