Maximizing Rotational Motion Energy

[preluderacer]

Homework Statement

In the figure, two blocks, of masses 2 kg and 3 kg, are connected by a light string which passes over a pulley of moment of inertia 4.00 E-3 kg m m and radius 0.05 m. The coefficient of friction for the table top is 0.30. The blocks are released from rest. Using energy methods, one can calculate that after the upper block has moved 0.6 m, its speed is:

The Attempt at a Solution

I don't even know where to start with this problem. A little guidance will be appreciated.

 

[tiny-tim]

hi preluderacer!

use conservation of energy (KE + PE = constant),

remembering that the speeds of the two blocks, and of the rim of the pulley, will be the same …

what do you get? ​

 

[preluderacer]

where does the friction force come into the equation?

 

[kjohnson]

It shows up as energy lost. So E1=E2+friction (where E1 & E2 are mechanical energies). And how would you figure out how much energy was lost to friction over that distance?

 

[preluderacer]

what i got is this (1/2)(I)(v/r)^2=(1/2)mv^2 - friction?

 

[tiny-tim]

hi preluderacer!

(try using the X² icon just above the Reply box )

what i got is this (1/2)(I)(v/r)^2=(1/2)mv^2 - friction?

no, not friction but work done by friction (and don't forget gravity)

 

[preluderacer]

im so confused is it then (1/2)mv²=(1/2)I(a/v)²+mu(d)+mgh?

 

[tiny-tim]

im so confused is it then (1/2)mv²=(1/2)I(a/v)²+mu(d)+mgh?

(have a mu: µ and an omega: ω )

yes, except two items are on the wrong side, aren't they?

and ω isn't a/v, and your µ is missing an N, and d = … ?

 

[preluderacer]

mgh+Nµ+(1/2)mv^2=(1/2)I(v/r)^2? I am so I am so confuded

 

[tiny-tim]

just think logically …

should the final KE of the two objects be on the same side or opposite sides?

should increasing h increase or decrease the final KE?

should increasing µN increase or decrease the final KE?