Force and work required to lift a 5kg sack at constant speed

[Jim4592]

Homework Statement

A 5.00 kg sack of flour is lifted vertically at a constant speed of 3.50 m/s through a height of 15.0 m.

Homework Equations

a) How great a force is required?
b) How much work is done on the sack by the lifting force?
c) What becomes of this work?

The Attempt at a Solution

Originally i assumed that acceleration at constant speed is the same as the speed, but then i realized that acceleration at constant speed is zero.

F=ma
F=5.00kg * 3.50 m/s
F = 17.5 kg * m/s
so when you include zero in the equation you get that there is no force?

But there has to be some force in order to calculate work

W=F*S*Cos(Θ)
W=F*(15.0m)*Cos(180) ; it would be 180 since its being lifted vertically correct?

 

[Jim4592]

I think i figured it out right after i typed it... would you use gravity acceleration?

F = 5 kg * -9.8 m/s
F = -49 kg m/s

W = -49 kg m/s * 15.0m * cos(180)
W = 735 J

Does that seem correct?

 

[PhanthomJay]

I think i figured it out right after i typed it... would you use gravity acceleration?

F = 5 kg * -9.8 m/s
F = -49 kg m/s

W = -49 kg m/s * 15.0m * cos(180)
W = 735 J

Does that seem correct?

You have the correct answer for the work, but your reasoning is a bit off. Since the object weighs 49N acting down (a kg m/s^2 is a Newton, by definition; the acceleration of gravity is 9.8 m/s^2 down), then in order for there to be no acceleration (constant speed implies no acceleration), the lifting force must be 49N up, (to give a net force of 0). The work done by that force is positive, since the displacement and lifting force are in the same direction. You seem to be calculating the work done by gravity, which is -735J (gravity acts down, displacement is up, so work done by gravity is negative.