# Constant Velocity = positive net force? But how to write it?

## 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 lets 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.

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SammyS
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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 lets 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.
You are apparently quite confused about several concepts. For one, acceleration is not force. For another, moving at constant velocity does not mean v/t = 0. It does mean that dv/dt = 0, i.e. the acceleration is zero.

Your notation also borders on being incomprehensible. Whatever do you mean by:
[W=F*d] = [W = MA*d] = [W = M(v/t)*d]

[W = M(0)*d] = [W = 0*d] = [W = 0]​
?

Question rewritten:

## 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] = [W = (M((d/t)/t))*d] and at a constant velocity: the rate of d/t=0 would yield:

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

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

Then I realize that maybe I am getting confused between lb_mass and lb_force

So my question is, what is the value of acceleration (or how is it expressed) in the formula [F = (mass)((distance/time)/time)] when a mass is being lifted at a constant velocity (on earth)?

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, and I'm just holding a bag of sand in a stationary position, then would the force be expressed as [F = (20lb_mass)(g)], and the Work expressed as [W=20lb_mass(g)*d = 0]?

If so, then if the bag of sand is being lifted at some constant velocity, how do I express the force of this in a formula? Would it be something like: F= (mass)(g+(some constant 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.

wait wait, i think i got it. It would be the same in either case! Since there is no acceleration in either case, the force on the stationary bag is the same as the force on the the bag with a constant velocity. F=(mass)(9.8) and the only way that would change is during the times of speeding up and slowing down, in which case it would just be [F=M(g+(some more acceleration)]

SammyS
Staff Emeritus