Tension and Work Calculations for Lifting a 285kg Load 22m Vertically

[fraxinus]

Homework Statement

A 285 kg load is lifted 22 m vertically with an acceleration a=.16 g by a single cable. Determine (a) the tension in the cable, (b) the net work done on the load, (c) the work done by the cable on the load, (d) the work done by gravity on the load, and (e) the final speed of the load assuming it started from rest.

Homework Equations

F=ma
Ft=g+ma?

The Attempt at a Solution

t= g + ma
t= g + (285)(-a)
t-mg= sigma F
t-mg=9.8-285a
t-2793= 9.8- 285a

 

[rock.freak667]

The Attempt at a Solution

t= g + ma
t= g + (285)(-a)
t-mg= sigma F
t-mg=9.8-285a
t-2793= 9.8- 285a

your equation should be ma=T-mg, so T=ma+mg. You can get T now.

 

[fraxinus]

so sigma F = T-mg? did you define y positive as up, and therefore tension as positive and mg negative? or am i mixed up?

 

[rock.freak667]

so sigma F = T-mg? did you define y positive as up, and therefore tension as positive and mg negative? or am i mixed up?

Yes I took up as positive.

 

[fraxinus]

sweet thanks! i got the right answer, but I'm still conceptually not understanding why we needed to both multiply and add the gravity to the acceleration. in other words, i never would have guessed to multiply .16 by 9.8... how do i know to do that? i know the g stands for gravity, but i don't see how i was supposed to know that.

now, for part b, i understand the formula for net work is W(net)= .5(m)(v)^2 - .5(m)(v)^2. however, when i simply multiply all the variables in the question together, i get the right answer. 285x22x9.8x.16= 9830 N. i don't want to just get the answers, i want to understand where they come from; am i supposed to use a velocity formula in here somewhere to get to the answer? thanks again!