Physics homework assist?

Question

Block 2 shown below slides along a frictionless table as block 1 falls. Both blocks are attached by a frictionless pulley. Find the speed of the blocks (in m/s) after they have each moved 3.7 m. Assume that they start at rest and that the pulley has negligible mass. Use 
m1 = 3.6 kg
 and
m2 = 7.2 kg.

oldschool · Accepted Answer

Acceleration = Fnet/(m1+m2) = 3.6*9.8/(3.6+7.2) = 49/15 m/s²
Vf² - Vi² = 2*a*d = Vf² - 0² = 2*(49/15)*3.7 = 4.92² 
Vf = 4.92m/s or 4.9m/s with 2 s.d.

? · Answer

loss of GPE: 3.6kg * 9.8m/s² * 3.7m = 131 J
and so that's the combined KE of both masses, which have the same speed:
131 J = ½ * (3.6 + 7.2)kg * v² → v = 4.9 m/s ◄
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Fireman · Answer

Let the tension in the string is T Newton and the acceleration of the system is a m/s^2, by balancing the net force on each block;
=>3.6a = 3.6g - T --------------(i)
& 7.2a = T ---------------------(ii)
By (i) + (ii):-
=>10.8a = 3.6g
=>a = (3.6 x 9.8)/10.8
=>a = 3.27 m/s
Let the velocity of each block is v m/s after each moved 3.7 m, Now by v^2 = u^2 + 2as
=>v^2 = 0 + 2 x 3.27 x 3.7
=>v = √24.17
=>v = 4.92 m/s

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Physics homework assist?

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