How much work is done on the ice block?

AI Thread Summary
The discussion focuses on calculating the work done on an ice block being pushed along an embankment. The user initially attempts to find the work using the formula W=Fxd, calculating the magnitudes of the displacement and force vectors. However, it is pointed out that the direction of the force relative to the displacement must also be considered in the work calculation. The user expresses uncertainty about their setup, indicating a need for clarification on the definition of work in this context. Understanding the directional components of both force and displacement is crucial for accurate work calculation.
Richober
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Homework Statement


An ice block is being pushed through a displacement d→=(15m)i -(12m)j along an embankment with rushing water. It exerts F→=(210N)i -(150N)J on block. How much work on block does force do?


Homework Equations


W=Fxd
Pythogoreum equation
a^2 +b^2= c^2



The Attempt at a Solution


I drew i and j coordinates for both displacement vector and force vector. I calculated
d→=15^2 + 12^2= c^2
d→=c=19.21m
I used same for F→
(210N)^2 + (150N)^2 = c^2
c=258.07 N
W=Fxd
(258.07N)* (19.21m) = 4957.52 J
I think I might have set it up wrong.
 
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Richober said:

Homework Statement


An ice block is being pushed through a displacement d→=(15m)i -(12m)j along an embankment with rushing water. It exerts F→=(210N)i -(150N)J on block. How much work on block does force do?

Homework Equations


W=Fxd
Pythogoreum equation
a^2 +b^2= c^2

The Attempt at a Solution


I drew i and j coordinates for both displacement vector and force vector. I calculated
d→=15^2 + 12^2= c^2
d→=c=19.21m
I used same for F→
(210N)^2 + (150N)^2 = c^2
c=258.07 N
W=Fxd
(258.07N)* (19.21m) = 4957.52 J
I think I might have set it up wrong.
Hello Richober. Welcome to PF !

What is the definition of work?

Not only are force, ##\ \vec{F}\,,\ ## and displacement, ##\ \vec{d}\,,\ ##involved, their relative directions must also be included..
 
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