# Coefficient of friction?

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A block of weight #6.1N# slides down a slope inclined at #tan^-1(11/60)# to the horizontal. The coefficient of friction between the block and the slope is #1/4# .

The block passes through a point A with speed #2m*s^-1# . Find how far the block moves from A before it comes to rest.

A block of weight

The block passes through a point A with speed

##### 1 Answer

Here's what I got.

#### Explanation:

The trick here is to realize that the block is actually *slowing down* as it's moving on the ramp.

In other words, the block is launched with an unknown initial velocity at the top of the ramp and it is **decelerating** as it's making its way down the ramp. This implies that the frictional force will *overpower* the projection of the block's weight along the length of the ramp.

Now, the forces that act on the block on the inclined plane can be represented like this

In your case, the **net force** that is acting on the block,

This implies that you have

#overbrace(F_"net")^(color(blue)("up the ramp")) = overbrace(" "f" ")^(color(blue)("up the ramp")) - overbrace(G * sin(theta))^(color(blue)("down the ramp"))" "color(blue)("(*)")#

Here

#f# is the frictional force#G = m * g# is theweightof the block, with#g# being the gravitational acceleration, which I'll take as#g = "9.8 m s"^(-2)# #theta# is the angle the ramp makes with the horizontal, in your case equal to#tan^(-1)(11/60)#

By definition, the frictional force is equal to

#f = mu * N#

Here

#mu# is the coefficient of friction between the block and the ramp#N# is thenormal force

As you can see in the diagram, you have

#N = G * cos(theta)#

which implies that

#f = mu * G * cos(theta)#

Plug this into equation

#F_"net" = mu * G * cos(theta) - G * sin(theta)#

Now, if you take *mass* of the block, you can say that

#overbrace(color(red)(cancel(color(black)(m))) * a)^(color(blue)(= "F"_ "net")) = mu * color(red)(cancel(color(black)(m))) * g * cos(theta) - color(red)(cancel(color(black)(m))) * g * sin(theta)#

which gets you

#a = g * [mucos(theta) - sin(theta)]" "color(blue)("(* *)")#

At this point, focus on finding the values of

You know that

#tan^(-1)(11/60) = theta#

so you can say that

#tan(theta) = 11/60#

Since

#tan(theta) = sin(theta)/cos(theta)#

you can say that

#sin(theta)/cos(theta) = 11/60#

This is equivalent to

#sin^2(theta)/cos^2(theta) = (11/60)^2#

and

#sin^2(theta) * 60^2 = cos^2(theta) * 11^2#

Use the fact that

#sin^2(theta) + cos^2(theta) = 1#

to rewrite this as

#[1 - cos^2(theta)] * 60^2 = cos^2(theta) * 11^2#

You should end up with

#{(cos(theta) = sqrt(60^2/(11^2 + 60^2)) = 60/61), (sin(theta) = sqrt(11^2/(11^2 + 60^2)) = 11/61) :}#

Now, you know that when the block passes through a point

If you take into account that the block will **stop** at a distance *after* it passes through point

#0^2 = v_"A"^2 - 2 * a * d#

Rearrange to solve for

#v_"A"^2 = 2 * a * d implies d = v_"A"^2/(2 * a)#

Finally, use equation

#d = (2^2 "m"^color(red)(cancel(color(black)(2)))color(red)(cancel(color(black)("s"^(-2)))))/( 2 * 9.8 color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-2)))) * (1/4 * 60/61 - 11/61))#

#d = color(darkgreen)(ul(color(black)("3.1 m")))#

I'll leave the answer rounded to two **sig figs**, but keep in mind that you only have one significant figure for the velocity of the block as it passes through point