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
AI Thread 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 ;)
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top