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°.