EK 1001 Questions

[Ehwic]

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335: Assume the force F acts for a time t. Assume the block has an inital velocity v. Initial velocity in which direction would result in the greatest mount of work done on the block?

A) To the right
B) to the left
c) initial velocity in either direction would result in the same amount of work done because the time t is constant
d) inital velocity in either direction would result in the same amount of work done because the distances traveled would be the same
answer: A

Why is it A? I realize W= fdcos0, however, does degree 0 really matter in this case? If it had gone to the left, wouldn't work still be the same, except just negative in value?

 

[ingramw1202]

335: Assume the force F acts for a time t. Assume the block has an inital velocity v. Initial velocity in which direction would result in the greatest mount of work done on the block?

A) To the right
B) to the left
c) initial velocity in either direction would result in the same amount of work done because the time t is constant
d) inital velocity in either direction would result in the same amount of work done because the distances traveled would be the same
answer: A

Why is it A? I realize W= fdcos0, however, does degree 0 really matter in this case? If it had gone to the left, wouldn't work still be the same, except just negative in value?

W= Fxd, where work is done when the force is applied in the same direction of motion. If a force F acts for time t, then I am assuming that F is pointing to the right.
W=KE=1/2(mv^2)
F=ma
Integral of acceleration function gives velocity function. If the velocity is in the same direction as the force, then the object accelerates. If the velocity is in the opposite direction of the applied force, then the object decelerates.
The problem doesn't give you a distance, simply a time. If the velocity and force are in the same direction, then for a given time, distance will be greater than if the two are in opposite directions.
Think of a car moving to the right for 10 seconds with an initial velocity of 20 m/s. If some force acts on the car to the right creating an acceleration, for a certain amount of time, then the total distance traveled would be:

x= vt + 0.5at^2, where vt is a positive number.

If the velocity is in the opposite direction of the force and consequently the acceleration, then:

x= -(vt) + 0.5at^2, where 0.5at^2 is the same for both situations and velocity is equal in magnitude but opposite in direction.

So as you can see, in the second situation, you would end up with a value for delta x that is less than the first value. And since W=Fxd, then W of the first one is greater than W of the second one because force is constant but distance is decreased.

 

[411309]

this is a very poorly worded question. I would of said it was the same too.

 

[ingramw1202]

this is a very poorly worded question. I would of said it was the same too.

I agree. Are they trying to keep us out of med school bahaha?

 

[Ehwic]

W= Fxd, where work is done when the force is applied in the same direction of motion. If a force F acts for time t, then I am assuming that F is pointing to the right.
W=KE=1/2(mv^2)
F=ma
Integral of acceleration function gives velocity function. If the velocity is in the same direction as the force, then the object accelerates. If the velocity is in the opposite direction of the applied force, then the object decelerates.
The problem doesn't give you a distance, simply a time. If the velocity and force are in the same direction, then for a given time, distance will be greater than if the two are in opposite directions.
Think of a car moving to the right for 10 seconds with an initial velocity of 20 m/s. If some force acts on the car to the right creating an acceleration, for a certain amount of time, then the total distance traveled would be:

x= vt + 0.5at^2, where vt is a positive number.

If the velocity is in the opposite direction of the force and consequently the acceleration, then:

x= -(vt) + 0.5at^2, where 0.5at^2 is the same for both situations and velocity is equal in magnitude but opposite in direction.

So as you can see, in the second situation, you would end up with a value for delta x that is less than the first value. And since W=Fxd, then W of the first one is greater than W of the second one because force is constant but distance is decreased.

Wow I gotta say, what a dick move made by EK

 

[ingramw1202]

Hahaha agreed!