Energy and WET (Incline pulley system)

AI Thread Summary
The discussion focuses on calculating the change in kinetic energy of a 45.0 kg block connected to a 105.0 kg block via a frictionless pulley system. The participants emphasize the importance of considering all forces, including string tension, when applying work-energy principles. There is confusion regarding the correct expression for potential energy and the role of tension, with suggestions to use energy conservation for the entire system. The calculations involve determining both the potential gain and the work done against friction. Ultimately, the need for clarity in applying the work-energy theorem and understanding the contributions of each force is highlighted.
closer
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A m1 = 45.0 kg block and a m2 = 105.0 kg block are connected by a string as in Figure P7.44. The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between the 45.0 kg block and incline is 0.250. Determine the change in the kinetic energy of the 45.0 kg block as it moves from A to B, a distance of 20.0 m.

7-44alt.gif


\DeltaKE = Wnet
\DeltaKE = Wg - Wfk
\DeltaKE = mgh - (uk)(N)(x)
\DeltaKE = mgh - (uk)(mgcos\theta)(x)

Final answer is incorrect. Any ideas? Thanks in advance.
 
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Maybe switch the trig and not trig around?

Gain in energy = potential gain: 20*sin(37)*9.8*45
loss in energy = friction work: (M1 Normal)*uk*20
 
closer said:
\DeltaKE = Wnet
\DeltaKE = Wg - Wfk
\DeltaKE = mgh - (uk)(N)(x)
\DeltaKE = mgh - (uk)(mgcos\theta)(x)
If you are going to apply the WET to m1 separately, you must include the work done by all forces, including string tension.

Also: Express h in terms of x. Be careful of the sign of the work contributions.
 
(delta KE) = (m2)(g)(h) - (m1)(g)(h)(sin(theta)) - (m)(g)(cos(theta))(uk)(x)

I added in the potential energy for the second block. Still incorrect. I don't understand why I would need tension, nor would I know how to incorporate it into the formula.

Edit: Isn't (m2)(g)(h) the string tension?
 
closer said:
(delta KE) = (m2)(g)(h) - (m1)(g)(h)(sin(theta)) - (m)(g)(cos(theta))(uk)(x)

I added in the potential energy for the second block. Still incorrect. I don't understand why I would need tension, nor would I know how to incorporate it into the formula.
If you are attempting to apply the WET to mass 1, then you need to include the work done by all forces on mass 1, including the string tension. Of course, this is the hard way to do this problem.

Edit: Isn't (m2)(g)(h) the string tension?
No.

I recommend that you use energy conservation on the system as a whole:
Ei + Wf = Ef

Where E is the total mechanical energy of both masses.
 
Maybe switch the trig and not trig around?

Gain in energy = potential gain: 20*sin(37)*9.8*45
loss in energy = friction work: (M1 Normal)*uk*20

Oops! Gain in energy = potential gain + kinetic gain; you would need to know initial and final kinetic for mine to work, I think. :frown:
 
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Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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