Grade 11 Energy Transformation with External Forces

AI Thread Summary
The discussion revolves around calculating the force exerted by a playground slide on a child sliding down. The child, weighing 15 kg, slides down a 4.0 m slide at a 40-degree angle, reaching a speed of 3.2 m/s at the bottom. The initial potential energy is considered, despite the child starting from rest, leading to confusion about energy transformations. The participant initially miscalculated the force of friction but later clarified their approach by incorporating gravitational potential energy into their calculations. Ultimately, they successfully determined the force exerted by the slide to be 75 N directed up the slide.
Wajeeha
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1. Question.

A 15 kg child slides from rest, down a playground slide that is 4.0m long. The slide makes a 40degree angle with the horizontal. The child's speed at the bottom is 3.2 m/s. What was the force that the slide was exerting on the child?
Answer: 75 N [up the slide]

2. My Attempt.

Ek = W = Efinal - Einitial
*Einitial cancels out because the child was at rest right? So its 0m/s
= 1/2(15kg)(3.2m/s)squared
= 76.8 J

Then I did:

W = Ff(costheta)(deltad)
Ff= W/(costheta)(deltad)
= 76.8 J/(cos180)(4)
= -19.2 N

3. Confusion:

I'm confused as to what I or like what my next step is? Could you explain what I might've done wrong or what I have to do after this and why? Maybe explain me the concept behind it?

 
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Wajeeha said:
Ek = W = Efinal - Einitial
*Einitial cancels out because the child was at rest right? So its 0m/s

The child may have been at rest, but are you sure that the child did not have any potential energy?
 
AlephNumbers said:
The child may have been at rest, but are you sure that the child did not have any potential energy?

Oh yeah! I did Eg = mgh; I found the haight using sinelaw then plugged that in. Then I found the Kinetic energy and subtracted that from the Eg which was the W then used the equation W = Ff(costheta)(deltad) and that's how the I got the Force of Friction. Thanks!
 
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