Answer: 105 N
Explanation:
The initial gravitational potential energy of the child is given by:
[tex]E_i = U_i = mgh_i = (25 kg)(9.8 m/s^2)(3.0 m)=735 J[/tex]
The child slides down at constant speed, so we there is no change in kinetic energy (because the speed remains constant). However, there is a change in gravitational potential energy, because when the child arrives at the bottom of the slide, the height is h=0 and the gravitational potential energy is Uf=0. Therefore, the variation of energy is
[tex]\Delta E= \Delta U=0-735 J=-735 J[/tex]
So, there have been a loss of energy. This energy loss is equal to the work done by the friction, which is given by:
[tex]W=-Fd[/tex]
where
F is the magnitude of the force of friction
d=7.0 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
Substituting, we get:
[tex]-735 J=-Fd\\F=\frac{-735 J}{-7.0 m}=105 N[/tex]
The magnitude of the kinetic friction force acting on the child is 105 Newton.
Given the following data:
We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 [tex]m/s^2[/tex]
To find the magnitude of the kinetic friction force acting on the child, we would apply the law of conservation of energy:
First of all, we would determine gravitational potential energy possessed by the child initially:
[tex]G.P.E_i = mgh\\\\G.P.E_i = 25 \times 9.8 \times 3[/tex]
G.P.Ei = 735 Joules
At the bottom of the slide, gravitational potential energy possessed by the child would be equal to zero (0) Joules.
So, the change in energy is given by:
[tex]\Delta E = 0 - G.P.E_i\\\\\Delta E = 0 - 735\\\\\Delta E = -735\;Joules[/tex]
The work done is the energy required to move the child from the top to the bottom of the slide.
[tex]Work\;done = \Delta E=-735[/tex]
Since friction opposes the direction, the work done is negative:
[tex]Workdone = -Fd\\\\-735 = -F(7)\\\\7F = 735\\\\F = \frac{735}{7}[/tex]
Kinetic friction force, F = 105 Newton
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