Question

A 100 kg crate is being pushed across a horizontal floor by a force P that makes an angle of 30.0o below the horizontal. The coefficient of kinetic friction is 0.200.What should be the magnitude of P, so that the net work done by it and the kinetic frictional force is zero?

Answer 1

Work =0
So Pcos30 -0.2*[100*9.8+Psin30]=0

or, 0.866P-196-0.1P=0
P = 255.875 N

Answer 2

First let's find the horizontal component of P say P' by considering negative fork done by friction f an positive work done by force P'

W_f=-μNd

W_p' = P'dwhere d is the distance traveled by crate before stopping.

since they are equal

W_p' +W_f =0 or

P'd -μNd =0

P'=μN where N=mg - Psin(θ)

The term -Psin(θ) is there since the force P is acting on the crate with its vertical component trying to lift the crate

now

P'=μ(mg- Psin(θ) )

and we know that P'= Pcos(θ) then

Pcos(θ) =μ(mg- Psin(θ) )

and finally

Pcos(θ) =μ(mg- Psin(θ) )

P(cos(θ) +μ sin(θ) ) =μmg

P=μmg/(cos(θ) +μ sin(θ) )

substituting

P=0.200 x 100 x 9.81/(cos(30) + 0.200 sin(30) )

P=203 N

Please let me know if you have any questions.