Confusion about work done by friction as negative or positive

[kjamha]

https://lh3.googleusercontent.com/0GOyZeQMdpLgYxYwwP4FMabe-Mk7QpADPC4ZK4PVpBGKvamcqjopOf3M0evdhE8PHZ3iTQ=s138 m1 is 4 kg and m2 is 8 kg. The kinetic friction coefficient on the table is 0.3. m1 is held in place. When m1 is released, m2 accelerates 1.2 m to the floor. Use conservation of energy/ and or Work KE principle to find the speed of the two blocks at the instant m2 hits the floor.

I used the following equation to solve the problem:

PE = KE1 + KE2 + work(friction) with g = 10 m/s2

8x10x1.2 = 1/2 (4) (v2) + 1/2 (8) (v2) + Nx.3x1.2 and then solved for v.

I got the right answer but I see work done by friction as negative (the force and displacement are in opposite directions). If I use a negative number here, I get the wrong answer. Also, regarding conservation of energy, I see a change in PE as negative energy (decrease in PE), the friction as negative work and the change in KE for m1 and m2 as a gain in energy. Shouldn't the change in PE + work done by friction = change in KE of m1 + change in KE of m2? Why is work done by friction positive in my original equation?

 

[Andreas C]

You've got the sign of Wf wrong. How did you arive at this equation? Because when non conservative forces are present, ΔΕ=W, where W is the work of the non conservative forces. ΔΕ=K1+K2-P0, so P0=K1+K2-Wf.

 

[Andreas C]

Another method giving the same result is this:
ΔΚ=Wf+Wg
Wg=-ΔP=P0
P0=K1+K2-Wf

 

[kjamha]

Your equation makes sense to me. Plugging in numbers for P0=K1+K2-Wf (using g = 10m/s2):
96 = 4v2 + 2v2 - 0.03x40x1.2
96 + 14.4 = 6v2
v = 4.29 m/s
If friction is eliminated, then 96 = 4v2 + 2v2 and v = 4 m/s. This is slower than 4.29. Adding friction should slow down the speed. That is why added Wf, which gives me the correct answer. I like your equations, but why is it giving me the wrong answer?

 

[Andreas C]

P0+Wf=K (where K=K1+K2), and since friction does a negative work, if you remove it, K is larger.

 

[kjamha]

I see it. Thanks!