Work Calculation for Box Pushed & Lifted

[mintyyf]

Work--uniform speed?

A 40.0 kg box is pushed 3.00m at uniform speed across a horizontal garage floor. and its then lifted 1 m into the back of a truck. Assuming that the force of friction acting between the box and the floor is 80.0 N what is the total work done in moving the box?



F = ma
W= Fd
Ffriction= Ufnet (u= coiffencet of friction)



So basically I started off by F = ma in which case accelration is = to zero because its going at a uniform speed now if that's the case then
W = O(F)3(d)
so that means work is zero as well. but then i was like how does friction work into that? i suppose its just to confuse me. Becuase what ever number i get at the beginning i'd add to the W = MGH
w = 40 x 9.81 x1
= 392 J
but if it the answer to the first part is zero wouldn't the total work be just the 392 from lifting it to the truck?? Hopefully that made sense and yu can help me clear this up thanks!

 

[PhanthomJay]

A 40.0 kg box is pushed 3.00m at uniform speed across a horizontal garage floor. and its then lifted 1 m into the back of a truck. Assuming that the force of friction acting between the box and the floor is 80.0 N what is the total work done in moving the box?



F = ma
W= Fd
Ffriction= Ufnet (u= coiffencet of friction)



So basically I started off by F = ma in which case accelration is = to zero because its going at a uniform speed now if that's the case then
W = O(F)3(d)
so that means work is zero as well. but then i was like how does friction work into that? i suppose its just to confuse me. Becuase what ever number i get at the beginning i'd add to the W = MGH
w = 40 x 9.81 x1
= 392 J
but if it the answer to the first part is zero wouldn't the total work be just the 392 from lifting it to the truck?? Hopefully that made sense and yu can help me clear this up thanks!

Yeah, well the total work done by all forces is zero, but the problem is asking, albeit not too clearly, what is the work done by the person pushing, then lifting, the block?

 

[mintyyf]

wouldn't the work just be a total of 392J being done by the person then? or am i missin something. this problem makes me feel extremely dense =(

 

[PhanthomJay]

wouldn't the work just be a total of 392J being done by the person then? or am i missin something. this problem makes me feel extremely dense =(

Yeah, you're missing something. Now when the person lifted the block , presumably at constant speed, you had no problem calculating her force against gravity, and the work done by her. What force does she apply against friction?

 

[mintyyf]

well that would be the ffrction = U x fnet
fnet =0
ffric = 80
which to gives me coffienct of friction...and that just furthur confuses me
i typed up the question as is. so i don't think i left out any pecies of information...i'm just so frustrated

 

[PhanthomJay]

well that would be the ffrction = U x fnet
fnet =0
ffric = 80
which to gives me coffienct of friction...and that just furthur confuses me
i typed up the question as is. so i don't think i left out any pecies of information...i'm just so frustrated

You had noted correctly that fnet=0. Since ffric = 80N, then with what force must the person push in order that fnet=0?

 

[HallsofIvy]

There is no coefficient of friction- you are given the actual force, 80N, so you don't need that.
Work= Force times Distance.

Of course, in lifting the box 1 m, you must overcome the force of gravity on it: 40g= 40(9.81). Again the work done is Force times Distance.