Mechanics problem: projectile experiencing friction

[James10492]

Hi there,

I have just come across this problem in my studies of mechanics and I am stuck for an answer:
Here is the set-up: a ball of mass 2.5kg is projected 2m away from the edge of a shelf 1.2m above the ground. The horizontal motion has a constant velocity of 5ms-1 (no acceleration). The surface is rough, so the ball experiences a force of friction as it traverses the 2m. When it leaves the surface it acts freely under the force of gravity and falls to the ground below. It takes 1s for the ball to complete its journey from the top of the shelf to the ground.

Part a) of the problem is to work out the amount of time the ball is in contact with the rough surface.

Resolving vertically from the point where the ball drops off the edge: initial speed downwards = 0, acceleration = g (9.8ms-2), and the displacement is 1.2m.

Using the constant acceleration formula, 1.2 = .5 * 9.8 * t^2; thus t = .494s. You know that the whole journey took 1s, so 1-.494 = .506s.


Now here is the part where I am very stuck. b) "Find the coefficient of friction between P and the surface".

Now if the surface was smooth you would expect the ball would only have been in contact with the surface .4s. So the force of friction has made it take .106s longer.

Normally I would use F = ma to find a resultant force and use that to work out the coefficient of friction (the force of friction itself being the normal reaction of the particle multiplied by the coefficient of friction (mu), so 2.5*9.8*mu). However the problem specifies that the ball is projected with constant velocity, no acceleration, so I do not know how to work out the resultant force. How am I supposed to work out the coefficient of friction?

 

[blamocur]

Looks to me like you are on the right track here. For the horizontal motion you know a) the initial speed, b) the total time and c) the travel time. Can you figure out the acceleration caused by the friction?

 

[James10492]

Ok I got it now.

Resolve horizontally in the same direction as the force of friction:

initial velocity = -5 displacement = -2 time = 5.1s

-2 = (-5*5.1) + (.5 * 5.1^2 * acceleration)

lets you deduce that the acceleration of the force of friction is 4.23.

F = ma, so F = 4.23 * 2.5 = 10.575

we know that F also = normal reaction * mu, so

10.575 = 2.5g * mu

mu = 10.575/2.5g = .432 (the answer in the book is actually .42 but I am satisfied)

 

[lex]

If you use 0.506 for your time (instead of the 5.1 you quote) you will get a more accurate answer.

Btw - in your problem statement, it must be that the initial velocity is 5m/s (not 'constant' during the horizontal motion, since friction is acting).