Leaving Cert Physics Question

Cian_

Hi guys, just need a bit of help with a question if you can help.

A bartender slides a pint of beer, of mass 600g along a bar, giving it an initial velocity of 6m/s. Given that the pint stops directly in front of a customer 7.2m away, what is the magnitude of the frictional force acting upon the glass?

Would be great if you could help me out.

Hint: use F = ma

You can calculate a (the acceleration, though in this case it's actually deceleration) from the information given, and you know m, so their sum will equal F. F in this case is F_friction — the magnitude of the frictional force. At least, that's how I imagine you'd do it without delving into slightly more advanced applied mathematics.

Cian_

gvn wrote: »

Hint: use F = ma

You can calculate a (the acceleration, though in this case it's actually deceleration) from the information given, and you know m, so their sum will equal F. F in this case is F_friction — the magnitude of the frictional force. At least, that's how I imagine you'd do it without delving into slightly more advanced applied mathematics.

I do Applied Mathematics so I tend to over think these types of physics questions and get a bit confused which is why I was asking for help lol, thanks for your help.

citrus burst

Here's an alternate way to do it, but may be a bit to advanced for you, it uses calculus.

p = m.v where p is momentum

F = dp/dt

so, F = d(mv)/dt = m dv/dt = m (v2-v1)/(t2-t1) you know v1, v2 and t1 so you need to find t2

x = 0.5(v1 + v2)t => t = 2 x/(v1 + v2)

Might work

An alternative would be to use:

v2^2 = v1^2 + 2ax

to find acceleration

It should be pretty trivial after that, just make sure to get your sign right!

FISMA

Cian_ wrote: »

Hi guys, just need a bit of help with a question if you can help.

A bartender slides a pint of beer, of mass 600g along a bar, giving it an initial velocity of 6m/s. Given that the pint stops directly in front of a customer 7.2m away, what is the magnitude of the frictional force acting upon the glass?

Would be great if you could help me out.

Cian,
Why did the pint stop? Clearly, there was an unbalanced force. Since this force acted over a distance, it did work on the pint.

The work done by friction stopped you, that is, it changed your kinetic energy.

Now just write the above sentence in what I call Phynglish and you're set.

Kirol

Cian_ wrote: »

Hi guys, just need a bit of help with a question if you can help.

A bartender slides a pint of beer, of mass 600g along a bar, giving it an initial velocity of 6m/s. Given that the pint stops directly in front of a customer 7.2m away, what is the magnitude of the frictional force acting upon the glass?

Would be great if you could help me out.

You know the pint stops so the final velocity v= 0 m/-2
Using v^2 = u^2 +2as (make v = 0)
And you get a= 2.5 m/s^2
F= ma. F= (.6kg)(2.5) F= 1.5N