# Conservation of energy problem with friction included

• simphys
In summary, 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.
simphys
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##

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.

simphys
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...

I mean c'mon now...

they'd be better off stating more generally ##\Delta K = W_{net}##

## What is the conservation of energy problem with friction included?

The conservation of energy problem with friction included refers to the principle that energy cannot be created or destroyed, but can only be transferred or converted into different forms. In the presence of friction, some of the energy in a system is converted into heat and lost, leading to a decrease in the total energy of the system.

## How does friction affect the conservation of energy?

Friction is a force that opposes motion and causes energy to be converted into heat. This means that in a system with friction, some of the initial energy will be lost and not conserved. This can result in a decrease in the total energy of the system over time.

## What is the role of friction in real-life conservation of energy problems?

In real-life conservation of energy problems, friction is a common factor that must be taken into account. For example, when calculating the energy required for a car to travel a certain distance, friction from the road and air resistance will decrease the overall energy efficiency of the car.

## How can friction be minimized in conservation of energy problems?

To minimize the effects of friction in conservation of energy problems, lubricants can be used to reduce the amount of friction between surfaces. Additionally, using smoother surfaces or reducing the weight of objects can also help to decrease friction and conserve more energy in a system.

## What are some examples of conservation of energy problems with friction included?

Some common examples of conservation of energy problems with friction included include calculating the energy required for a car to travel a certain distance, determining the efficiency of a roller coaster, and analyzing the energy transfer in a pendulum system. In all of these examples, friction plays a role in decreasing the total energy of the system.

• Introductory Physics Homework Help
Replies
28
Views
1K
• Introductory Physics Homework Help
Replies
20
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
997
• Introductory Physics Homework Help
Replies
41
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
769
• Mechanics
Replies
2
Views
907
• Introductory Physics Homework Help
Replies
4
Views
431
• Introductory Physics Homework Help
Replies
12
Views
1K
• Introductory Physics Homework Help
Replies
15
Views
339
• Introductory Physics Homework Help
Replies
8
Views
898