<p>Hi I’m having trouble with these problems…I’d really appareciate it if someone could explain them to me. thanks in advance.</p>
<li><p>An object slides on a level surface in the +x direction. It slows and comes to a stop with a constant acceleration of -2.45 m/s^2. What is the coefficient of kinetic friction between the object and the floor?</p></li>
<li><p>A horizontal force of 5.0 N accelerates a 4.0 kg mass, from rest, at a rate of 0.5 m/s^2 in the positive direction. What friction force acts on the mass?</p></li>
<li><p>A student pulls a box of books on a smooth horizontal floor with a force of 100 N in a direction 37 degrees above the horizontal. If the mass of the box and books is 40 kg, what is the acceleration of the box?</p></li>
</ol>
<p>oy, ap physics was last year.. do you have the equations page? I could help if you have the equation that has the coefficient of kinetic friction in it</p>
<p>force of friction = normal force * coefficient of kinetic/static friction</p>
<p>i'll work those out, give me a minute...</p>
<p>1) do you know the mass? if you did you could do summation of forces in the x direction, which in this case appears to be only friction. so -f = -ma. you know a and m so you can solve for f. and since f = (mu)(N) you can set those two equal to eachother. (and N = mg because it's the same as the weight force.)</p>
<p>2) summation of forces in x dir = 5 - f = 4(.5) now solve for f.</p>
<p>don't have time right now to explain #3. good luck!</p>
<ol>
<li>A student pulls a box of books on a smooth horizontal floor with a force of 100 N in a direction 37 degrees above the horizontal. If the mass of the box and books is 40 kg, what is the acceleration of the box?</li>
</ol>
<p>there is two way to solve this problem and i assume you want a simpler one (which ignore weight lifted by vertical force.</p>
<p>cos 37 x 100N / 40kg
horizontal-only force / mass
meter/sec^2*kg / kg</p>
<p>These are really the only equations you need:
Fr = Fn * Mu
Fn = mg
F = ma</p>
<p>F=force in N
Fr=frictional force in N
Fn=normal force in N
Mu=coefficient of friction (no unit)
m=mass in kg
g=acceleration due to gravity, 9.81 m/s^2
a=acceleration in m/s^2</p>
<ol>
<li>
Don't know, sorry. There's probably some way to cancel out the variables you don't know, ie mass, force...</li>
</ol>
<p>2.
5 - Fr = 4 * .5
5 - Fr = 2
Fr = 3 N</p>
<ol>
<li>Assuming negligible friction,
force along the x-axis is 100cos(37) = 79.86 N
F=ma
79.86 = 40a
a = 1.997 m/s^2</li>
</ol>
<p>thanks a lot guys, I really appreciate all of your help. This board is filled with so many bright people =)</p>
<ol>
<li>An object slides on a level surface in the +x direction. It slows and comes to a stop with a constant acceleration of -2.45 m/s^2. What is the coefficient of kinetic friction between the object and the floor?</li>
</ol>
<p>4 this one, you dont have to know "mass"</p>
<p>frictional force = frictional acc. x m = (mu) x mg --- divide both sides by m
frictional acc. = mu x gravity
-2.45 m/s^2 = mu x 9.8 m/s^2
mu = -2.45/9.8</p>
<p>thanks dreaming, I would have never figured that out. My teacher is terrible, he talks to the board the whole time and then expects to understand everything at the end of his lectures.</p>
<p>sorry..just one more question if its not too much trouble. I don't understand how you incorporate the kinetic friction into this problem to find the acceleration.</p>
<p>An object with a mass of 12 kg slides down a rough 37 degree inclined plane, where the coefficient of kinetic friction is 0.20. What is the acceleration of the object?</p>
<p>You can figure this one out on your own - the only way that this is different from, say, #2 above is that it's going down an inclined plane.</p>
<p>The key equation here is still Fr = Fn * Mu, then after that, use F = ma</p>
<p>Draw out a diagram of this problem before you do it. You will need to use a little bit of trig to find Fn, but after that just plug in the values and solve.</p>
<p>ok..so I get..
Fn = (.012)(9.81 m/s^2) = .118
Fr = (.118)(.2) = 2.35x10^-2
2.35x10^-2 = (.012)a
a = 1.962 m/s^2</p>
<p>Not quite. Try drawing this out, and label ALL forces that interact in this problem. Draw an x-y axis and observe which forces are going in which direction. </p>
<p>With an incline plane, your normal force won't be just mass * gravity, because it's applied at an angle</p>
<p>Also, F = ma, in this case, would be better expressed as Fgravityx - Ffriction</p>
<p>(Sorry if this isnt very coherent [or accurate], I'm kinda sleepy...)</p>