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a 31 kg child slides down a playground slide at a constant speed. the slide has a height of 3.0 m and is 7.8 m long. part a using the law of conservation of energy, find the magnitude of the kinetic friction force acting on the child.

Sagot :

116 N

The initial  potential energy of the child is given by:

[tex]U_{i} =mgh[/tex]

[tex]U_{i} =31[/tex]×[tex]9.8[/tex]×3 = 911.4 J

There is no change in kinetic energy since the child is sliding down at a steady speed (because the speed remains constant). When the child reaches the bottom of the slide, the height is h=0 and the gravitational potential energy is [tex]U_{f}[/tex]=0, hence there is a change in the gravitational potential energy. Consequently, the energy variation is

Δ[tex]E[/tex]=ΔU= 0-911.4 = -911.4J

So, there have been a loss of energy. This energy loss is equal to the work done by the friction, which is given by:

W=-Fd

where F is the magnitude of the force of friction, d=7.8 m is the length of the slide ,the negative sign is due to the fact that the friction is opposite to the motion of the child

On Substituting, we get

-911.4=-Fd

∴F= [tex]\frac{911.4}{7.8}[/tex]

F=116N

To learn more about Kinetic Friction here :https://brainly.com/question/11574957

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