Two blocks are connected by a very light string passing over a massless and frictionless pulley (Figure (Figure 1) ). The 20.0-N block moves 75.0 cm to the right and the 12.0-N block moves 75.0 cm downward.
Question: Find the total work done on 20.0-N block if there is no friction between the table and the 20.0-N block.
ΣF = ma
Wtotal = F * s cos(φ), where the force is constant.
Wtotal = 1/2(m)vf^2 - 1/2(m)vi^2
vf^2 = vi^2 + 2ad
The Attempt at a Solution
So I started with the 12.0N weight.
ΣF = m1a = T - 12.0N, where m1 is the mass of the 12.0N weight. I also say acceleration is the same for both weights because they are in the same system.
then T = (12/9.8)a+12.0N
Then I set up my force equation for the 20.0N weight.
ΣF = m2a = T, where m2 = 20/9.8
(20/9.8)a = (12/9.8)a + 12.0N
(8/9.8)a = 12.0N
a = 14.7 m/s^2
Then I plug this in to my equation for tension.
T = (12/9.8)14.7 + 12.0
Then W = T * s cos(0°)
W = (30)(.75)(1) = 22.5 J.
Which is wrong.
I thought maybe force is not constant, although I'm not sure why it wouldn't be. SO i used this equation.
W = 1/2(m)vf^2 - 1/2(m)vi^2, where i assume vi = 0m/s
W = 1/2(m)vf^2
to solve for vf: vf^2 = vi^2 +2ad
vf^2 = 2(14.7)(.75)
vf^2 = 22.05
W = 1/2(20)(22.05)
W = 220.5J
which is also wrong. I know my displacement vector is correct so it must be my acceleration but I can't see anywhere that i made an error in calculating it.