# Force of Friction (1 Viewer)

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#### msimard8

This is my problem

A 15 kg child slides, from rest, down a playground slide tht is 4.0 m long, as shown in the figure. The slide makes a 40 degrees angle with the horizontal. The child's speed at the bottom is 3.2 m/s. What was teh force of friction that the sldie was exerting on the child?

Ok. The answer in the texbook says 75N.

Here is my WROng attempt

work=Kinetic Energy 2 - Kinetic Energy 1
work = 1/2 m (V2squared - v1squared)
w=1/2 (15 kg) (3.2 m/s squared - 0)
w=7.5 (10.24 m/s)
w=76.8 J

then I..

Work=force (distance) (cos40)
f= w/d (cos 40)
f=76.8 J / 4.0 (cos 40)
f=14.7 N

Where is my error.. HELP PLEASE

#### renatzu

msimard8 said:
work=Kinetic Energy 2 - Kinetic Energy 1
work = 1/2 m (V2squared - v1squared)

w=1/2 (15 kg) (3.2 m/s squared - 0)
w=7.5 (10.24 m/s)
w=76.8 J
then I..
Work=force (distance) (cos40)
f= w/d (cos 40)
f=76.8 J / 4.0 (cos 40)
f=14.7 N
Where is my error.. HELP PLEASE
I've bolded the problem. Kinetic energy is not conserved, total energy is. That means that you must take potential energy into account.

#### msimard8

I took what you said into effect. I still dont get 75 N as a result.

Given

m=15 kg
g=9.81m/s squared
final velocity=7.2 m/s
initial velocity = 0m/s
d=4.0 m

w= gravitational potential energy 1 - gravitational potential energy 2
w= [(m) (g) (h2)] - [(m) (g) (h1)]
w= mg (h2-h1)
w= 15 kg (9.81 m/s squared) (4.0m - 0)
w= 588.6 J

then

work = force (distance) (cos 40)
f=w/d cos 40
f= 588.6 J / 4.0 m (0.7660)
f= 588.6 J / 3.0641
f=192.1

therefore the answer is still wrong according to the textbook

How does Velocity relate to this problem?

Last edited:

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