coefficient of kinetic friction?

Question

Objects of masses m1 = 4.00 kg and m2 = 9.00 kg are connected by a light string that passes over a frictionless pulley . The object m1 is held at rest on the floor, and m2 rests on a fixed incline of = 40.0°. The objects are released from rest, and m2 slides 1.00 m down the incline in 4.00 s.

a) Determine the coefficient of friction between m2 and the incline.

b) determine the tension in the string

Answer 1

It is an interesting question since it involves a total force that changes as the system is sent in motion. The distance from the center of the pulley to the center of block of m2 is missing. Let’s call it H. As the mass m2 moves the string will make, an increasing with motion, angle A. [at the start angle A=0]. Since it we are looking at a friction force we have to consider forces normal and parallel to the inclined plane.

Tension on the rope is caused by a force that moves block m1 up, m2 down and does that against friction f.

Summation of normal forces
1. Fn= N2-N1 where
N2= m2 g cos(40)
N1= m1g sin(50-A) wonder how I got 50 – A? (if not then get back to me. Hint:( 90-A)-40))
Where A= arctan( S cos(40) / H) this is why we need the distance H from the center of the pulley to the center of block of m2
Summation of parallel is responsible for systems downward motion.
2. Fp= F2-F1-f
F2= m2g sin(40)
F1= m1f
f= uFn = u[= m2 g cos(40) - m1g sin(50-A) ]

The force that produces the acceleration is Fp=ma or
3. Fp=a(m1+m2)
Just be careful acceleration ‘a’ is not a constant acceleration.

That should get you started on the right track.

Answer 2

Hey, there, it is a fairly good question, but also too much to talk about, I can tell you some hints to do it. First off, you need a free-body diagram, it is extreamly helpful in this question, I did. Then, you will notice that the top one on the floor is simpler to figure out, considering horizontally, the tension force minus the friction force is the sum to cause the movement of both objects, and you know f=the coefficient * FN, and from the energy conservation, find out the final speed at 4s later, thus you will know the acceleration. Over all th process, you will have a expression of T, then bring it to the inclined one, use free-body diagram agian to analyze what caused the this object to move, list the equation you got, and replace the T into what you got in the first step, then you will have a expression of coefficient of all the numbers.

Answer 3

The coefficient of kinetic friction is the ratio between the kinetic friction tension and the conventional tension performing on the article. This ratio is regularly a relentless selection reckoning on the form of fabrics in touch. Kinetic friction is that incredibly resisting tension which varies with the conventional tension performing on the article. they selection linearly, meaning that if the conventional tension doubles, kinetic friction doubles, however the ratio will stay a relentless selection. *Edit to characteristic: Kinetic friction will stay a relentless cost of F = (coefficient of kinetic friction) * N, so as which will fairly be the optimal friction. With static friction on the different hand, it tiers from 0 to (coefficient of static friction)*N.

Answer 4

First lets calculate the acceleration of m2

s=1/2at²+ut
1=1/2(a)(4)²+(0)(4)
a=0.125 m/s²

Now lets calculate the friction force .
F=ma

m2gsinx - Friction - Tension = m2a
Tension = m1a
(9)(9.8)sin40 - Friction - 4(0.125)=9(0.125)
Friction = 57.31886717 N

Friction=µR
R=m2gcosx
57.31886717 =µ(9)(9.8)cos40
µ=0.8483499662

b) Tension = m1a
Tension = 4(0.125)=0.5 N