What is the optimal speed for a box to slide down a roof with minimal friction?

[enrijuan]

Homework Statement

You and your friend Peter are putting new shingles on a roof pitched at 20^$$\circ$$ . You're sitting on the very top of the roof when Peter, who is at the edge of the roof directly below you, 5.1 m away, asks you for the box of nails. Rather than carry the 2.1 kg box of nails down to Peter, you decide to give the box a push and have it slide down to him.

Homework Equations

If the coefficient of kinetic friction between the box and the roof is 0.54, with what speed should you push the box to have it gently come to rest right at the edge of the roof?

The Attempt at a Solution

I thought about conservation of mechanical energy
W=$$\Delta$$KE + PE
mu(9.81cos20)(5.1/sin20)= -.5v_i² + (9.81)(5.1)

I got the wrong initial velocity can anyone help?

 

[JesseC]

Energy is not conserved in this example because friction is what's known as a non-conservative force. As a general rule of thumb, whenever friction is involved, energy will not be conserved.

Do you know how to draw a free body diagram?

You need to think about the different forces acting on the box (there are only two!) and in what direction they are acting.

From this you can work out the resultant force that the box experiences, and therefore the deceleration it experiences. When you know the deceleration, use the equation of motion ('SUVAT' equation) to find the initial velocity.

 

[tiny-tim]

hi enrijuan!

You and your friend Peter are putting new shingles on a roof pitched at 20^$$\circ$$ . You're sitting on the very top of the roof when Peter, who is at the edge of the roof directly below you, 5.1 m away, asks you for the box of nails. Rather than carry the 2.1 kg box of nails down to Peter, you decide to give the box a push and have it slide down to him.
…
I thought about conservation of mechanical energy
W=$$\Delta$$KE + PE
mu(9.81cos20)(5.1/sin20)= -.5v_i² + (9.81)(5.1)

this is fine , except you're confused about the distances …

(did you draw a clear diagram for yourself? … always do that, to avoid mistakes)

both your (5.1/sin20) and your (5.1) are wrong ​

(btw, you mustn't call this "conservation of mechanical energy" …

it's the work-energy theorem …

as JesseC says, energy isn't conserved when work is done by friction )

 

[enrijuan]

I thought that there would be force due to gravity, normal force and the force due to kinetic friction? and I drew a diagram and thought that the distance was directly below as in the y-axis and the hypotenuse would be the displacement.

 

[tiny-tim]

… I drew a diagram and thought that the distance was directly below as in the y-axis and the hypotenuse would be the displacement.

ah, now i see what's been confusing you …

… You're sitting on the very top of the roof when Peter, who is at the edge of the roof directly below you, 5.1 m away …

yup, "directly below" would normally mean as in the y-axis, but in the context of a sloping roof, it must mean "directly downhill", so the 5.1 given in the question is measured down the slope, not vertically

so yes, the hypotenuse is the displacement, but the question (not very clearly) says that it's 5.1

 

[enrijuan]

ah ok so then my original equation is correct? as in

W= $$\Delta$$KE + PE ?

mu(9.81cos20)(5.1)= -.5v_i² + (9.81)(5.1sin20) ?

or should I find the Net Force as JesseC said? If I take that route then
F_x=f-F_gx right?
F_x=mu(m)(g)-m(g)(sin20)
F_x=.54(2.1)(9.81cos20)-2.1(9.81sin20)= 3.41 N
F_x= 3.41= ma
a= -1.62 m/s^2

S=5.1
U= trying to find
V=0
A= -1.62 m/s^2
T= don't know

so, use v^2=vi^2+2as

0=vi^2 + 2(-1.62)(5.1)
v_i=(2*1.62*5.1)^.5

v_i=4.06 m/s is that right?

 

[tiny-tim]

ah ok so then my original equation is correct? as in

W= $$\Delta$$KE + PE ?

mu(9.81cos20)(5.1)= -.5v_i² + (9.81)(5.1sin20) ?

(have a mu: µ and a delta: ∆ )

yes that's correct, except i think the LHS should have a minus

(JesseC's method should give the same result, but it's much longer … using the work-energy theorem is a short-cut which effectively eliminates an integration of the equations of motion)