Messages No.3741 от Jake 10 April 2008 18:25
Theme: HELP!!! Masses on a pulley

Hi, I got this question in an online assesment and got it 'wrong'. I can't see where I went wrong but maybe I'm missing something.

Two blocks with different mass are attached to either end of a light rope that passes over a light, frictionless pulley that is suspended from the ceiling. The masses are released from rest, and the more massive one starts to descend. After this block has descended a distance 1.40m , its speed is 2.00m/s . If the mass of the 2 blocks combined is 13kg, what is the mass of the larger block?

I can only seem to get 7kg, but the answer says it should be 7.45.

• 4186: Re: HELP!!! Masses on a pulley drg 28 September 2008 21:08
  

So a=(2*2)/(2*1.4)=2/1.4 m/s²

Use Newton's Second Law

∑F_ext= (m₁+m₂)a
You know that m₁+m₂=13
so M_total*a=13*(2/1.4)
but ∑F_ext= m₂g-m₁g
if m₂ is the larger block, which is pulling down.

Two unknowns, two equations.
(1) m₁+m₂=13
(2) g(m₂-m₁)=13*(2/1.4)
Solve it and you'll get m₂=7.45 rounded up ;)

• 4247: Re: HELP!!! Masses on a pulley Andrea de Capoa 04 марта 19:29
  

> Two blocks with different mass are attached to either end of a light rope that passes over a light, frictionless pulley that is suspended from the ceiling. The masses are released from rest, and the more massive one starts to descend. After this block has descended a distance 1.40m , its speed is 2.00m/s . If the mass of the 2 blocks combined is 13kg, what is the mass of the larger block?

> I can only seem to get 7kg, but the answer says it should be 7.45.

Using the Energy conservation:
Mgh-mgh=0,5*(M+m)*v^2

You get
M-m = 1.9

And then
M=7.45

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