Question

Moving at an initial speed of vi = 2.00 m/s, Jimmy slides from a height of h = 5.00 m down a straight playground slide, which is inclined at θ = 55° above the horizontal. At the bottom of the slide, Jimmy is moving at vf = 9.00 m/s. By determining the change in mechanical energy, calculate the coefficient of kinetic friction between Jimmy and the slide.

Answer 1

Let mass of jimmy be m and the coeffiecient of kinetic friction be .

The forces acting on jimmy are weight, frictional force and normal force due to the incline.

Taking the component of forces along the incline and normal to the incline :-

There is no motion in the direction perpendicular to the incline, therefore acceleration will be 0 i.e normal force and component of weight perpendicular to the incline(mg cos 55⁰) will balance each other .

N = mg cos 55⁰

Frictional force = * N = * mg cos 55⁰

Here the vertical height is 5 m.Therefore , the length of the incline will be .

We know from work energy theorem that total work done is equal to change in kinetic energy .i.e

W_total = final kinetic energy - initial kinetic energy ____(1)

W_total = work done by gravity + work done by friction + work done by normal

= mg*5 - *  mg cos 55⁰*(length of incline) + 0 (normal force is perpedicular to velocity therfore work done by normal force is 0 also g is the acceleration due to gravity)

= mg*5 -   * mg cos 55⁰ *

= 5mg -   *5mg cot 55⁰

final kinetic energy =

initial kinetic energy =

Therefore , from 1

5mg -   * 5mg cot 55⁰ =

10g -   * 10g cot 55⁰​​​​​​​ = 77

10g(1- *cot 55⁰​​​​​​​ ) = 77 (g = 9.8m/s²)

On solving ,

= 0.306