Calculating Net Work: Solving for P with Coefficient of Kinetic Friction

  • Thread starter Thread starter Morokana
  • Start date Start date
  • Tags Tags
    Work Work done
Click For Summary
To solve for the force P required to achieve zero net work while pulling a 95.0-kg crate, one must resolve the forces into their x and y components, considering the coefficient of kinetic friction of 0.284. The net force in both the x and y directions must equal zero, indicating no acceleration. The frictional force can be calculated using the weight of the crate multiplied by the coefficient of kinetic friction. The relationship Pcos34.3 = Ffrict helps in determining the magnitude of P. Properly resolving the forces leads to the correct solution for P.
Morokana
Messages
6
Reaction score
0
A 95.0-kg crate is being pulled across a horizontal floor by a force P that makes an angle of 34.3° above the horizontal. The coefficient of kinetic friction is 0.284. What should be the magnitude of P, so that the net work done by it and the kinetic frictional force is zero?

i could not do this question .. i tried so much ! ... i tried finding out P.E. .. but wut do i do with the coefficient of kinetic friction ??... Work done = force X distance .. but wheres the distance .. please help me in this
 
Physics news on Phys.org
Man, we should have a compilation of Physics Forums Solved Problems in Physics :P We could even sell it one day.

Alright so basically this is a force resolving problem. You need to resolve all the forces into their respective x and y components, Then do a summation of the forces. The question has stated that there is no acceleration in the x direction (forces are equal, Fnet = ma, Fnet = 0, a = 0). Obviously there is no acceleration in the y direction as well (it's on the floor).

Fnetx = 0
Fnety = 0

The trick is resolving the forces into the x and y components. But it seems like all the angles are given. I say give it a try.
 
Thanx a lot man .. i got the answer right :) .. you do .. Fnet=95 x 9.8

then u multiply the answer by 0.284
 
Yup, then Pcos34.3 = Ffrict (in x).
So you didn't really need help after all ;)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
Replies
18
Views
3K
  • · Replies 58 ·
2
Replies
58
Views
5K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
29
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
8
Views
3K
  • · Replies 7 ·
Replies
7
Views
8K
Replies
1
Views
2K