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DUCKPIN BOWLING BALLS SET OF 3 YELLOW & BLACK MANHATTAN RUBBER |  |
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Rubber Bowling Ball Set
Can anyone help me understand Friction in Physics?
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?
I'm begging, i have a thesis this semester, and i have no time to answer this set of problems T_T please i'm begging...
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.
En lille tight bowlingkugle
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