Work Energy Theorem with Kinetic Friction and External Work

In summary: In this equation, "Work done by friction" represents the work done by friction, which is negative (since friction opposes the displacement).
  • #1
kc0ldeah
12
0

Homework Statement



A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.5 m/s. The pulling force is 100 N parallel to the incline which makes an angle of 20.0° with the horizontal. The coefficient of kinetic friction is .4 and the cart is pulled 5.00 m.

e) What is the speed of the crate after being pulled 5.00m?

Homework Equations


Delta Energy Mechanical = -Force_friction(d) + Work_external

K_f + U_f = K_i + U_i - Force_friction(d) + Work_external

For constant Force and parallel force and displacement: W = F(d)

The Attempt at a Solution



K_f + U_f = K_i + U_i - Force_friction(d) + Work_external

.5mvf2 + mgh = .5mvi2 + 0 - Force_friction(d) + Work_external

h = 5sin(20°)
Force_friction = -μ Fn
= -μ 98cos(20°)
Work External = F(d) = 100N * 5m = 500J

.5mvf2 + mgh = .5mvi2 +μ 98cos(20°)(5) + 500

vf2 = (.5mvi2 +μ 98cos(20°)(5) + 500 - mgh) / (.5m)

v = 10.27 m/s

The answer is given as 5.65 m/s.
Any help is appreciated!
 
Last edited:
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  • #2
kc0ldeah said:
vf2 = (.5mvi2 +μ 98sin(20°)(5) + 500 - mgh) / (.5m)
Is that a typo, or did you change cosine into sine?
 
  • #3
Sorry that was a typo I will fix it.
 
  • #4
kc0ldeah said:
.5mvf2 + mgh = .5mvi2 +μ 98cos(20°)(5) + 500
Check the sign of the work done by friction.
 
  • #5
Doc Al said:
Check the sign of the work done by friction.

K_f + U_f = K_i + U_i - Force_friction(d) + Work_external

If you plug -μ98cos(20°) into that it should be +μ98cos(20°) right?
It is -μ98cos(20°) to begin with because Force_friction acts in the opposite direction of the crate's movement.

EDIT: But fixing it the way you said gives me 5.65 which is the answer. Why is that term supposed to be negative?
 
  • #6
kc0ldeah said:
K_f + U_f = K_i + U_i - Force_friction(d) + Work_external

If you plug -μ98cos(20°) into that it should be +μ98cos(20°) right?
It is -μ98cos(20°) to begin with because Force_friction acts in the opposite direction of the crate's movement.

EDIT: But fixing it the way you said gives me 5.65 which is the answer. Why is that term supposed to be negative?
It depends on how you define the terms. That friction term represents the work done by friction, which is negative (since friction opposes the displacement). When you wrote "-Force_friction(d)", I assume you meant for Force_friction to be the magnitude of that force.

More generally: Work done by friction (or any force) = Force*distance = (-μmgcosθ)(d)

Final Energy = Initial Energy + Work done by friction + Work done by applied force
 
  • #7
Final Energy = Initial Energy + Work done by friction + Work done by applied force

Thank you this is what I really needed to see.

I guess the equation my teacher gave me already factored in the fact that friction is an opposing force. That has really been messing me up.
 
  • #8
kc0ldeah said:
I guess the equation my teacher gave me already factored in the fact that friction is an opposing force.
Exactly.
 

1. What is the Work Energy Theorem?

The Work Energy Theorem states that the work done on an object is equal to the change in its kinetic energy. This means that when a force is applied to an object and it moves a certain distance, the work done by that force results in a change in the object's kinetic energy.

2. How does Kinetic Friction affect the Work Energy Theorem?

Kinetic friction is a force that opposes the motion of an object. When considering the Work Energy Theorem, kinetic friction must be taken into account because it is a force that does work on the object, resulting in a decrease in its kinetic energy. This means that the work done by the applied force must be greater than the work done by kinetic friction in order to result in a net increase in kinetic energy.

3. What is External Work in relation to the Work Energy Theorem?

External work refers to the work done on an object by external forces, such as gravity or a person pushing or pulling the object. The Work Energy Theorem takes into account both external work and the work done by internal forces, such as friction, to determine the overall change in an object's kinetic energy.

4. How is the Work Energy Theorem used in real-world applications?

The Work Energy Theorem is used in various fields, such as physics, engineering, and mechanics, to analyze the motion and energy of objects. It is particularly useful in calculating the work done and the forces involved in simple machines, such as pulleys and levers. It is also used in analyzing the motion of objects in situations involving friction, such as car braking systems.

5. What are some limitations of the Work Energy Theorem?

The Work Energy Theorem is based on the assumption that all forces involved in the motion of an object are conservative forces. This means that the work done by these forces is independent of the path taken by the object. In real-world situations, there may be non-conservative forces, such as air resistance, that can affect the work done and the overall energy of the object. Additionally, the Work Energy Theorem does not take into account changes in potential energy, which may be significant in certain situations.

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