Can anyone help me understand Friction in Physics?

Question

My brain is really bleeding in Physics, i will failed in my Physics class if i will not do my assignment, can anyone help me answering this problems please.

1. The coefficient of sliding friction between a rubber tire and a wet concrete road is 0.5

a. find the minimum time in which a car whose initial velocity is 30 miles/hour can come to a stop on such a road.

b. what distance will that car cover in this time?

2. A bowling ball with an initial velocity of 3.0 m/s rolls along a level floor for 50m before coming to a stop. What is the coefficient of rolling friction?

3. An 80 labs wooden crate rests on a horizontal wooder floor. If the coefficient of static friction is 0.5, how much force is needed to set the crate in motion?

Answer 1

Wow mom's really laying it on thick.  She's got a good point, though. Whatever here you go:

1) First find the friction force slowing the car down:  

F = mu * m * g

Acceleration (deceleration, actually) from F = m * a

mu * m * g = m * a; solve for a:

a = mu * g = .5 * 9.8 = 4.9 m/s^2

then to find time use

Vf = Vo - a * t; solve for t with Vf = 0

t = Vo / a = 13.4 / 4.9 = 2.73 sec

Stopping distance from

X = Vo * t - 1/2 * a * t^2

X = 13.4 * 2.73 - 0.5 * 4.9 * (2.73)^2

X = 18.3 m

2) Sorry it's been awhile since 10th grade physics; brain hurts too much to remember rotational friction.

3) Pretty basic stuff. Force = 392 lbf. If you can't figure this one out you have no business in a physics class sorry!

Answer 2

are you asking someone to do your home work for you?

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Answer 3

1a)

Let M = mass of the tyre.

Considering horizontal road, the normal force on the car by the road =

Mg

Force of friction = coefficient of friction * normal force = 0.5Mg

Accceleration a = -friction/mass = -0.5Mg/M = -0.5 g (negative because friction is opposite to movement)

Or, a = -0.5 * 9.8 m/s^2 = -4.9 m/s^2

Initial velocity u = 30 miles/hour

1 mile = 1600 meter, 1 hour = 3600 sec

Therefore, u = 30 * 1600/3600 m/s = 13.33 m/s

Final velocity v = 0

Time t = ?

v = u + at

Or, 0 = 13.33 - 4.9* t

Or, t = 13.33/4.9 second = 2.72 second

b) Distance s = ut + 1/2 at^2

= 13.33 * 2.72 - 1/2 * 4.9 * 2.72^2 = 36.26 - 18.13 = 18.13 meter

2) Initial velocity u = 3.0 m/s

Final velocity v = 0

Displacement s = 50 m

Let a = acceleration

v^2 = u^2 + 2as

Or, a = (v^2 - u^2)/2s

Or, a = (0 - 3^2)/(2 * 50)

Or, a = -9/100 m/s^2

Or, a = -0.09 m/s^2

Negative sign shows that it is deceleration.

Deceleration = 0.09 m/s^2

Let f = force of friction

Then f = M * deceleration, where M = mass of the ball

Or, f = 0.09 M

Normal force on the ball by the floor = weight = Mg

Coefficient of rolling friction = force of friction/normal force

= 0.09/(Mg) = 0.09/g = 0.09/9.8 = 0.009

3) Mass m = 80 lbs

Acceleration due to gravity g = 32 ft/s^2

Weight = mg = 80 * 32 lb-ft/s^2 = 2560 lb-ft/s^2

Normal force on the crate by the floor = weight of the crate = 2560 lb-ft/s^2

Force of friction on the crate by the floor = coefficient of friction * normal force = 0.5 * 2560 lb-ft/s^2 = 1280 lb-ft/s^2

In order to set the crate in motion, force has to be applied to oppose friction.

Therefore, minimum 1280 lb-ft/s^2 force is needed.

I have answered all your questions. Contact me if you have any other doubts.