What is the energy lost due to friction?

AI Thread Summary
The discussion revolves around a physics problem involving a sledge loaded with bricks, where the user is trying to calculate the tension in a rope, work done by the rope, and energy lost due to friction. The user initially struggles with finding the correct tension due to not considering the vertical component of the tension affecting the normal force. After some guidance, they realize they need to use trigonometric functions, specifically tangent, to solve the problem correctly. Ultimately, the user successfully figures out the calculations needed for the assignment. The conversation highlights the importance of understanding the components of forces in physics problems.
tristan_fc
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Okay, I have this problem I've been fighting with for a couple of hours now. I have the answers for it, but I can't figure out how they're the answers. Here's the problem:

A sledge loaded with bricks has a total mass of 18.0 kg and is pulled at constant speed by a rope. The rope is inclined at 20° above the horizontal, and the sledge moves a distance of 20.0 m on a horizontal surface. The coefficient of kinetic friction between the sledge and surface is 0.500.

A) What is the tension of the rope?

B) How much work is done on the sledge by the rope?

c) What is the energy lost due to friction?

If I got the first answer right, I could do the rest, but that's where I'm stuck. I know I need to find the x-component of the tension, and then use trig to find the tension of the rope. What I did was this:

Tx = 0.5*18*9.8 = 88.2

That's wrong. I know it's wrong, because the tension I found with that value was approx. 82.9. The answer is supposed to be 79.4. I know it's wrong, because I'm not considering the y-component of tension, but I don't know how to do that. Anybody have any ideas? This is a web-based assignment, due by midnight central time. (4 hours)Thanks for your help.
 
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It's wrong because you forgot that the normal force is reduced by the vertical component of the tension.

Is that enough to get you started again?
 
Well that's what I said.

I figured it out though. I remembered that you have to use tangent. Thanks though. :)
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

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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