Conservation of energy problem with friction included

AI Thread Summary
The discussion revolves around a conservation of energy problem involving a crate being pulled with friction. The initial velocity of the crate is calculated as 10.458 m/s for the first 15 meters, where the surface is frictionless. The net force acting on the crate is determined to be 114.56 N after accounting for friction, leading to the equation for the final velocity. A confusion arises regarding the sign in the energy equation, with clarification that the work-energy theorem should be applied correctly to avoid errors. The conversation concludes with a suggestion to simplify the approach by focusing on total work done rather than intermediate velocities.
simphys
Messages
327
Reaction score
46
Homework Statement
A 96-kg crate, starting from rest, is pulled across a floor
with a constant horizontal force of 350 N. For the first 15 m
the floor is frictionless, and for the next 15 m the coefficient
of friction is 0.25. What is the final speed of the crate?
Relevant Equations
conservation of energy
so I haven't looked at the solution yet, but I know that a 100% the velocity needs to be bigger, but analytically, I get a - sign instead of a + sign as you'll see at the final square root.

So for the first 15meters of the motion all you should know is that ##v_1 = 10.458 m/s##.

for the 2nd part:
energy is conserved right. 2 NC-forces are acting on it but net force is positive however.
So ##E_1 = E_2 + W_{NC}## (1)
##W_{NC} = F{net,NC}*\Delta x##

to get ##F_{net,NC}## , we have ##F = 350N## and the friction force is found from an FBD
so ##F_r = \mu_kN = \mu_kmg##
##F_{net,NC} = F - \mu_kmg = 350 - 0.25*96*9.81 = 114.56N##
Now back to ##(1)##:
##\frac12mv_1^2 = \frac12mv_2^2 + F_{net,NC}\Delta x## where work done by the NC forces is pos.
rearranging to ##v_2##, we get:
##v_2 = \sqrt{\frac12mv_1^2 - F_{net,NC}\Delta x}## WHERE i GET A - SIGN INSTEAD OF A + SIGN HERE, WHY?
##v_2 = 8.6m/s##
 
Physics news on Phys.org
simphys said:
Homework Statement:: A 96-kg crate, starting from rest, is pulled across a floor
with a constant horizontal force of 350 N. For the first 15 m
the floor is frictionless, and for the next 15 m the coefficient
of friction is 0.25. What is the final speed of the crate?
Relevant Equations:: conservation of energy

WHY?
Because you put the term on the wrong side of the equation. The work-energy theorem states that the difference between kinetic energy after and before equals the work done.

However, you are needlessly complicating things here by computing the intermediate velocity. You have two forces, the pulling force and the friction force. The first is constant and acts over a total of 30 m. The second is also constant (once applied) and acts over a total of 15 m. You can easily compute the total work done from when the crate was at rest.
 
Orodruin said:
Because you put the term on the wrong side of the equation. The work-energy theorem states that the difference between kinetic energy after and before equals the work done.

However, you are needlessly complicating things here by conputing the intermediate velocity. You have two forces, the pulling force and the friction force. The first is constant and acts over a total of 30 m. The second is also constant (once applied) and acts over a total of 15 m. You can easily compute the total work done from when the crate was at rest.
Oh gosh, I just thought of it myself that it could've prob been done by considering like that, but then I thought nehh we have two different forces, but of course you are totally right, thank you!

And okay... thank you I wasn't really sure which equatoin the actual statement for conserv of E would be. That's why I stated that one aka ##E_1 = E_2 + W_{NC}##, so I'll just use ##\delta K = W_C + W_{NC}## from now on, thank you.
 
See @Orodruin it's soo confusing in this book...
1657795259321.png

I mean c'mon now...

they'd be better off stating more generally ##\Delta K = W_{net}##
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top