Calculating Work for a Pulled Carton Across Floor

  • Thread starter Thread starter DiscreteCharm
  • Start date Start date
  • Tags Tags
    Series
AI Thread Summary
To calculate the work done by a horizontal force pulling a 21 kg carton across the floor at constant speed, the applied force must equal the friction force due to the coefficient of kinetic friction (0.41). Since there is no acceleration, the net force is zero, indicating that the applied force balances the friction force. The normal force on a horizontal surface is determined by the equation N = mg, where m is the mass and g is the acceleration due to gravity. If the carton were on an incline, the normal force would be calculated based on the forces acting perpendicular to the incline. Understanding these concepts is crucial for solving similar physics problems effectively.
DiscreteCharm
Messages
3
Reaction score
0

Homework Statement



A horizontal applied force, F, pulls a 21 kg carton across the floor at constant speed. If the coefficient of kinetic friction between the carton and the floor is 0.41, how much work does the applied force do in moving the carton 2.5 m?

Homework Equations



I'm supposed to be working with energy techniques, we just covered work/power/energy conservation/and so on

The Attempt at a Solution



I don't quite understand, it seems to me like the force is 0 given there is no acceleration.
 
Physics news on Phys.org
DiscreteCharm said:
I don't quite understand, it seems to me like the force is 0 given there is no acceleration.
The fact that there is no acceleration tells you that the net force is zero. Hint: Figure out what the friction force must be and then deduce what F is.
 
That gives way to the problem I have been having in working out a few other problems. I know it's a simple question, but how do I determine normal force (since I'm almost always given the coefficient of kinetic friction)? Is it simply mass times gravity (on a horiztonal surface, at least)?
 
Yes, in this case, the normal force is simply N = mg. In general, you figure it out by setting the sum of the vertical forces equal to zero. The only vertical forces acting on the carton are -mg and +N, so N - mg = 0, and thus N = mg. (Sometimes the applied force will have a vertical component that will affect the normal force.)

And if the carton were being pushed up an incline instead of on a horizontal surface, you'd figure out the normal force by setting the sum of the forces perpendicular to the incline equal to zero.
 
Thank you very much, that was extremely helpful.
 
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