Question

An object of mass m is at rest at the top of a smooth slope of height h and length L. The coefficient of kinetic friction between the object and the surface, mk, is small enough that the object will slide down the slope if given a very small push to get it started. Find an expression for the object’s speed at the bottom of the slope.

Answer 1

wut?

Answer 2

find acceleration and integrate it? idk how I'm supposed to solve this problem lol. This is a physics problem btw.

Answer 3

@agent0smith

Answer 4

You don't need integration; acceleration will be constant.

First thing, as always, is a free body diagram. Then work out the net force in the y and x directions.

Answer 5

Like that?

Answer 6

I asummed the slope to be on the x axis

Answer 7

ohh ok

Answer 8

ok

Answer 9

What am I solving for? acceleration?

Answer 10

Yes, using net force equations

Answer 11

hehe \[F_{n}-mgcos \theta = m a \]

Answer 12

y components

Answer 13

Yes, but a is zero for y

Answer 14

\[F_{f}-mgsin \theta = ma\]

Answer 15

look

Answer 16

Ff = mu*Fn

Answer 17

\[a= \frac{ \mu_{k}F_{n}-mgsin \theta }{ m }\]

Answer 18

Use the y Fnet to solve for Fn.

Answer 19

Why do I want to solve for Fn?

Answer 20

Because you have an Fn.

Answer 21

\[F_{n}=\frac{ ma+mg \sin \theta }{ \mu_{k} }\]

Answer 22

"Yes, but a is zero for y"

Answer 23

Use the y-Fnet to solve for Fn, NOT the x-Fnet.

Answer 24

ok \[F_n=mgcos \theta\]

Answer 25

Now you can plug it into your a equation and simplify.

Answer 26

\[a= \frac{ \mu_{k}mgcos \theta - mgsin \theta }{ m }\]

Answer 27

Simplify. And keep in mind we don't know theta, but we have the height and length of the plane, so we can find cos and sine.

Answer 28

\[a=g( \mu_k \cos \theta- \sin \theta) \]

Answer 29

Find cos and sine by drawing a triangle, and then once you have them, use \[\Large v_f^2 = v_i^2 + 2a x\]

Answer 30

what is that?

Answer 31

sin theta = h/x cos theta= L/ x

Answer 32

ohh lol

Answer 33

sin theta= h/ L cos theta = x/L x= L cos theta

Answer 34

Use pythag.

Answer 35

\[\sqrt{L^2 - h^2}\]

Answer 36

Better.

Answer 37

How do you go about representing speed?

Answer 38

The equation I gave you earlier, vf is final speed. We can assume initial is ~zero.

Answer 39

Plug in your big ugly expression for a. x would be the distance which is L

Answer 40

\[v= \sqrt{2L[g(\mu_k \cos \theta-\sin \theta)}]\]

Answer 41

:s

Answer 42

Good but you forgot sine and cosine

Answer 43

lol so I just replace cos theta and sin theta with L/x and h/x respectively

Answer 44

There should be no x. That's why you used pythag...

Answer 45

And your sin theta is wrong. SOHCAHTOA.

Answer 46

My bad I was looking at my first notes which were wrong yes so I meant sin theta= h/L and cos theta= sqrt(L^2-h^2)

Answer 47

I'm so tired lol sorry

Answer 48

Now cos theta is wrong :P

Answer 49

Ahhhhhhh :s

Answer 50

cos theta= adj/hyp.... adj = sqrt(L^2-h^2)??

Answer 51

ohhh typo lol I forgot to put it over L

Answer 52

I hate this question for not just saying " and the angle of the slope is theta". Makes the final expression so much worse looking.

Answer 53

omg ikr

Answer 54

\[\large v= \sqrt{2Lg \left( \mu_k \frac{ \sqrt{L^2-h^2} }{ L }-\frac{ h }{ L }\right)} \]

Answer 55

D:

Answer 56

Then cancel the common L \[\large v= \sqrt{2g \left( \mu_k \sqrt{L^2-h^2} - h \right)}\]who the F decided that this should be the expression instead of using theta.

Answer 57

ooooohhh :}

Answer 58

Ohhh well there is a part B to this and you're given actual values for height and length. look

Answer 59

I can just plug in the values into the equation lol