Calculating Work Done for Carrying Tool Boxes

[cyberjupiter]

Homework Statement

A carpenter is holding two tool boxes of mass 19kg and 21kg respectively. Calculate the work done to carry the tool boxes through a horizontal distance of 16m.

Homework Equations

W = mgh

The Attempt at a Solution

m = 21kg + 19kg
g = 9.81ms^-2
h = 0

W = (21 + 19)(9.81)(0)
W = 0J

Is this correct? And if this is correct, would you explain why it is 0J?

 

[NascentOxygen]

The formula for work involves the dot product of two vectors.

W = F . s

Does this give you a hint?

(Yes, your answer of 0J is correct.)

 

[HallsofIvy]

In order to do work, you must move an object with mass in the direction of a force vector.

 

[cyberjupiter]

So if I use W = F . s, then it would be

W = (19 + 21)(9.81)(16) ?
Or is it,
W = (19 + 21)(9.81)(16) cos 90 ?

 

[NascentOxygen]

So if I use W = F . s, then it would be

W = (19 + 21)(9.81)(16) ?
Or is it,
W = (19 + 21)(9.81)(16) cos 90 ?

this ^^^

Instead of writing the vector formula, it is often expressed as W = F s cosθ
and that's probably the best one to memorise.

 

[cyberjupiter]

NascentOxygen, would you tell me why the angle is 90? Shouldnt it be 0? And how about if I use sin 0? I will still get 0J.

 

[NascentOxygen]

NascentOxygen, would you tell me why the angle is 90? Shouldnt it be 0? And how about if I use sin 0? I will still get 0J.

The question just tests how easily confused you are.

The only force mentioned is weight, and it acts downwards ⇓
while the motion described is horizontal ⇒
and the angle between those two vector quantities is ... ?

You want to invent your own formula using sin 0° ? What angle in this picture is 0°?