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

## 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

Ray Vickson
Homework Helper
Dearly Missed

## 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 whats 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

Chestermiller
Mentor
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

Chestermiller
Mentor
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

Chestermiller
Mentor
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

Chestermiller
Mentor
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 gotta 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