Calculate angle for least work done

[NewtonGalileo]

Homework Statement

A man wishes to pull a crate 15 m across a rough floor by exerting a force of 100 N. The coefficient of kinetic friction is 0.25. For the man to do the least work, the angle between the force and the horizontal should be:
0
14
43
66
76

Homework Equations

work done = f*d*cos(theta)
friction force = mu * normal force
mu = 0.25

The Attempt at a Solution

total work done = f*d*cos(theta) - mu*m*g*d = m*a*d = 0 (since a =0 to minimise work)

 

[a.ratnaparkhi]

If it is 76,then accept my opinion.
As angle increases, value of cos\theta decreases. So at maximum angle, work done will be least.As 76 is maximum in options, according to me it is the answer.

 

[tiny-tim]

Welcome to PF!

Hi NewtonGalileo! Welcome to PF!

(have a mu: µ and a theta: θ )

work done = f*d*cos(theta)
friction force = mu * normal force
mu = 0.25

The Attempt at a Solution

total work done = f*d*cos(theta) - mu*m*g*d = m*a*d = 0 (since a =0 to minimise work)

No, the normal force is not mg, is it?

However, the question asks "For the man to do the least work …", so I don't think the work done by the friction force matters.

 

[NewtonGalileo]

I thought 76 degrees also. But, the answer given in the answer key is 0 degrees. Does not make sense. Am I missing something?

 

[tiny-tim]

Hi NewtonGalileo!

(just got up :zzz: …)

I agree with you (and a.ratnaparkhi) …

the magnitude |F| of the force is fixed,

and the distance d pulled is fixed.

so the work done is F.d, = Fdcosθ, which is a minimum when θ is largest, ie 76°.