- #1

LearninDaMath

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## Homework Statement

Lifting a 20lb bucket of sand a distance of 100 ft at a rate of 2ft/s. I would assume:

[W=F*d] = [W = MA*d] = [W = M(v/t)*d] and at a constant velocity: rate of v/t=0 so

[W = M(0)*d] = [W = 0*d] = [W = 0]

However, a video online shows that you'd take 20lb*100 ft = 2000 ft lbs of work.

Then I realize I am getting confused between lb_mass and lb_force

So my question is, what is the value of acceleration in F = (mass)

__((distance/time)/time)__that let's F have a value?

If I'm holding a 20lb bag of sand and gravity is pulling down at g, does that just mean that I'm lifting a mass with an acceleration equal and opposite of the value of gravity? So if gravity is value -g, then if I'm just holding a bag of sand in a stationary position, then the force is [F = (20lb_mass)(g)], while the Work is [W=20lb_mass(g)*d = 0]

So if the bag of sand is being lifted at some constant velocity, how do I express the force of this: F= (mass)(g+(some velocity)? The acceleration of gravity doesn't change, its always 9.8m/s^2, but the upward lift now must be some value greater than the downward force of gravity. I'm confused about how the formula of F=MA at zero velocity differs from F=MA at

a constant velocity.