How much work is done on the ice block?

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SUMMARY

The discussion centers on calculating the work done on an ice block being pushed along a displacement vector of d→=(15m)i -(12m)j with a force vector of F→=(210N)i -(150N)j. The initial calculations yielded a displacement magnitude of 19.21m and a force magnitude of 258.07N, leading to an incorrect work calculation of 4957.52 J. The key takeaway is that the direction of the force relative to the displacement must be considered to accurately compute work.

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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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