# Challenge problem -- rock sliding up and over a roof into an arc....

• Ab17
In summary: Thank you for your helpIn summary, the roofer kicks a round, flat rock that slides up a 37.0° sloped roof with an initial speed of 15.0 m/s. With a coefficient of kinetic friction of 0.400, the rock slides 10.0 m up the roof before crossing the ridge and going into free fall with negligible air resistance. Using energy, the maximum height reached by the rock above the point where it was kicked is 6.84 m. When using Newton's laws, the final velocity is 6.66 m/s and the maximum height reached is 7.66 m.
Ab17

## Homework Statement

One side of the roof of a house slopes up at 37.0°. A roofer kicks a round, flat rock that has been thrown onto the roof by a neighborhood child. The rock slides straight up the incline with an initial speed of 15.0 m/s. The coefficient of kinetic friction between the rock and the roof is 0.400. The rock slides 10.0 m up the roof to its peak. It crosses the ridge and goes into free fall, following a parabolic trajectory above the far side of the roof, with negligible air resistance. Determine the maximum height the rock reaches above the point where it was kicked.

## The Attempt at a Solution

I know how to solve the problem with energy. I want to know how can you apply Newtons laws to get the solution

Ab17 said:

## Homework Statement

One side of the roof of a house slopes up at 37.0°. A roofer kicks a round, flat rock that has been thrown onto the roof by a neighborhood child. The rock slides straight up the incline with an initial speed of 15.0 m/s. The coefficient of kinetic friction between the rock and the roof is 0.400. The rock slides 10.0 m up the roof to its peak. It crosses the ridge and goes into free fall, following a parabolic trajectory above the far side of the roof, with negligible air resistance. Determine the maximum height the rock reaches above the point where it was kicked.

## The Attempt at a Solution

I know how to solve the problem with energy. I want to know how can you apply Newtons laws to get the solution

First: show us you work (PF rules).

If you use Newtons laws then both mgsin@ and fk will act down the incline and the block is moving up the incline while there is no force. So this is what's confusing me.

Using energy I done this:
Wf + Eo=Ef
(Ukmgcos@)dcos180 +1/2mvi2 = 1/2vf2 + mgh
Substituting values we get:
-(0.4)(9.8)(cos37)(10) + 0.5(15)^2 = (9.8)(10sin37) + 1/2vf2
Vf= 5.35
Then using projectile motion for Hmax

Let's see what your force balance equation looks like in the direction along the slope.

Just found this now:

F=ma
Mgsin@ -fk =ma
Mgsin@ -UkN = ma
Mgsin@ - Ukmgcos@ = ma
gsin@ - Ukgcos@ = a
Substituting values we get a = 5.57
Vf2= vi2+ 2ax
Vf =

Vf = 18.34

Ab17 said:
Just found this now:

F=ma
Mgsin@ -fk =ma
Mgsin@ -UkN = ma
Mgsin@ - Ukmgcos@ = ma
gsin@ - Ukgcos@ = a
Substituting values we get a = 5.57
These equations are incorrect. If a is the acceleration up the slope, then
$$-mg\sin\theta-\mu_kmgcos\theta=ma$$
So a is negative. Using both your energy equation and this equation, I get 6.67 m/s for the final velocity and -9.02 m/s^2 for a.

How can the object accelerate up the slope while there is no force applied in that direction

Ab17 said:
How can the object accelerate up the slope while there is no force applied in that direction
It's acceleration up the slope is negative, so its acceleration vector is pointing down the slope. There is a force applied down the slope. In fact there are two of them.

How do you get 6.67ms-1 from the energy eqn can you please show. Did I do something wrong there because I got 5.35

Ab17 said:
How do you get 6.67ms-1 from the energy eqn can you please show. Did I do something wrong there because I got 5.35
You just got to do the arithmetic correctly.

(0.4)(9.8)(cos37)(10)=31.31

(9.8)(10sin37)=58.98

0.5(15)^2=112.5

-31.31 +112.5 = 58.98 + 1/2vf2

81.19 - 58.98 = 1/2vf2

22.21 = 1/2vf2

##v_f^2=44.42##

##v_f=6.66##

Using this to find hmax we get 0.82m while the book gives the answer to be 6.84m

Ok I see its 10sin37 + the 0.82

## 1. How does the angle of the roof affect the rock sliding over it?

The angle of the roof can greatly impact the trajectory and speed of the rock as it slides over. A steeper angle will result in a faster and more direct descent, while a flatter angle may cause the rock to roll or bounce before reaching the arc.

## 2. What factors contribute to the rock being able to slide up and over the roof?

The main contributing factor is the initial force or energy given to the rock to start its movement. This can come from a variety of sources such as gravity, a push or throw, or even wind. Additionally, the shape and weight of the rock, as well as the material and texture of the roof, can also play a role.

## 3. Is it possible for a rock to slide over a roof and then fall back down?

Yes, it is possible for a rock to slide over a roof and then fall back down, especially if the initial force is not strong enough to carry it all the way to the arc. This can also happen if there are any obstacles or uneven surfaces on the roof that cause the rock to lose its momentum.

## 4. How does the arc affect the trajectory of the rock?

The arc acts as a barrier or obstacle for the rock, causing it to change direction and potentially lose some of its energy. The shape and angle of the arc can also greatly impact the final trajectory of the rock, as well as any other objects or surfaces it may come into contact with after sliding over the roof.

## 5. Can the challenge problem be applied to real-life scenarios?

Yes, this challenge problem can be applied to real-life scenarios, especially in the fields of physics and engineering. It can help scientists and engineers better understand the principles of motion and trajectory, and how different factors can affect the movement of objects. It can also be used in practical applications, such as designing structures to withstand potential impacts from sliding objects.

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