Force of Friction (1 Viewer)

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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
 
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.
 
still no correct answer

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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