Question

A steel playground slide is 5.25 m long and is raised 2.75 m on one end. A 45.0 kg child slides down from the top starting at rest. The final speed of the child at the bottom is 6.81 m/s. Find the average force of friction between the child and the slide.

Answer 1

Answer:

$F=32.24N$

Explanation:

From the question we are told that:

Height $h= 2.75 m$

Length$l = 5.25 m$

Mass $m=45kg$

Final speed $v_f=6.81$

Generally the equation for Potential Energy P.E is mathematically given by

$P.E=mgh$

Therefore

Initial potential energy

$P.E_1=45*9.8*2.75 \\\\P.E_1= 1212.75 J$

Generally the equation for Kinetic Energy K.E is mathematically given by

$K.E=0.5mv^2$

Therefore

Final kinetic energy

$K.E_2= 1/2*45*6.81*6.81 \\\\K.E_2= 1043.46J$

Generally the equation for Work_done is mathematically given by

$W=P.E_1-K.E_2\\\\W=169.3$

Therefore

$F=\frac{W}{d}\\\\F=\frac{169.3}{5.25}$

$F=32.24N$