A mass m1 = 4.8 kg rests on a frictionless table and connected by a massless string over a massless pulley to another mass m2 = 4.5 kg which hangs freely from the string. When released, the hanging mass falls a distance d = 0.77 m.

Question

idku · Answer

I can't figure out:

(Part 3)   What is the final speed of the two blocks?
(Part 4)   How much work is done by tension on m1?

idku · Answer

$\color{#000000}{    \displaystyle   F=ma      }$

$\color{#000000}{    \displaystyle   F(\Delta t)^2=m\Delta x      }$

$\color{#000000}{    \displaystyle   F(\Delta t)^2=(m_1+m_2)d      }$

the force here is the weight of m2,

$\color{#000000}{    \displaystyle   (m_2g)(\Delta t)^2=(m_1+m_2)d      }$

$\color{#000000}{    \displaystyle   (9.80*4.5)(\Delta t)^2=(4.8+4.5)*0.77      }$

$\Delta t=0.280360$

$\color{#000000}{    \displaystyle   F=ma      }$

$\color{#000000}{    \displaystyle   F(\Delta t)=(\Delta v)m      }$

$\color{#000000}{    \displaystyle   (9.80*4.5)*(0.402965)=(v_f-0)(4.8+4.5)      }$

$\color{#000000}{    \displaystyle   v_f=1.91083      }$

idku · Answer

this is what I attempted but turned out to be incorrect.