Work Problem: Spring and Friction, find final Speed

• JoeyBob
In summary, the conversation revolved around calculating non-conservative work from friction, which was found to be -8.82 using the formula Ff=umg. The initial kinetic energy was given as 136.89 and the change in potential energy was determined to be 8.1216. The equation Ekf-Eki+Change Ep=Work NC was used to calculate the final kinetic energy, which resulted in a value of 119.9484. The participants also discussed using units in calculations and the importance of keeping an open mind when using variables for multiple purposes.
JoeyBob
Homework Statement
See attached
Relevant Equations
Ek=1/2mv^2 Ep=1/2 k (x)^2
First calculated non conservative work from friction using Ff=umg. Non conservative work was -8.82.

Initial kinetic energy, 1/2mv^2, was 136.89.

Change in potential energy, 1/2k(x)^2, was 8.1216.

Ekf-Eki+Change Ep=Work NC

Ekf=W NC+Eki-change Ep

=-8.82+136.89-8.1216=119.9484

Ekf=1/2mv^2, v=7.3014 but answer should be 7.44. I've confirmed there were no rounding errors by trying the same calculation all at once in the calculator (I still get 7.3).

Attachments

• Question.PNG
7.8 KB · Views: 87

TSny said:
Wouldnt it just be the friction force here, umg? This is 0.2*9.8*4.5=8.82. Friction force would be against it so it would also be negative.

JoeyBob said:
Wouldnt it just be the friction force here, umg? This is 0.2*9.8*4.5=8.82. Friction force would be against it so it would also be negative.
Force and work do not have the same dimensions. So, the work cannot equal the force.

JoeyBob
TSny said:
Force and work do not have the same dimensions. So, the work cannot equal the force.
So I use the distance its been compressed to find the work. Thanks.

JoeyBob said:
So I use the distance its been compressed to find the work. Thanks.
OK. Good. Note that you would have probably caught your error if you had included units in your calculations.

TSny said:
OK. Good. Note that you would have probably caught your error if you had included units in your calculations.
Its more that I wasnt open minded enough to use that variable for two things - I only saw it as part of the potential energy of the spring (the compression).

1. What is the equation for calculating the final speed in a spring and friction work problem?

The equation for calculating the final speed in a spring and friction work problem is vf = sqrt((2kx^2)/m - (2ukx^2)/m), where vf is the final speed, k is the spring constant, x is the displacement of the spring, m is the mass of the object, and u is the coefficient of friction.

2. How do you determine the direction of the final speed in a spring and friction work problem?

The direction of the final speed can be determined by looking at the direction of the net force acting on the object. If the net force is in the same direction as the displacement, the final speed will be in the same direction. If the net force is in the opposite direction, the final speed will be in the opposite direction.

3. What is the role of friction in a spring and friction work problem?

Friction plays a crucial role in a spring and friction work problem as it is the force that opposes the motion of the object and causes it to slow down. Without friction, the object would continue to move at a constant speed after being released from the spring.

4. How does the coefficient of friction affect the final speed in a spring and friction work problem?

The coefficient of friction directly affects the final speed in a spring and friction work problem. A higher coefficient of friction means there is more resistance to the motion of the object, resulting in a lower final speed. A lower coefficient of friction means there is less resistance, resulting in a higher final speed.

5. Can the final speed in a spring and friction work problem be greater than the initial speed?

Yes, the final speed in a spring and friction work problem can be greater than the initial speed. This can occur if the spring has a high enough spring constant and the coefficient of friction is low enough, allowing the object to gain more speed as it moves through the frictional force.

• Introductory Physics Homework Help
Replies
2
Views
245
• Introductory Physics Homework Help
Replies
21
Views
646
• Introductory Physics Homework Help
Replies
3
Views
667
• Introductory Physics Homework Help
Replies
21
Views
2K
• Introductory Physics Homework Help
Replies
8
Views
952
• Introductory Physics Homework Help
Replies
15
Views
1K
• Introductory Physics Homework Help
Replies
8
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
8
Views
2K
• Introductory Physics Homework Help
Replies
11
Views
1K