What Are Common Physics Homework Questions on Work, Energy, and Vectors?

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Common physics homework questions on work, energy, and vectors often involve concepts such as power, kinetic energy, and the effects of forces. The derivative of work with respect to time is identified as power, while the dot product of two vectors being zero indicates an angle of 90°. Kinetic energy is dependent on an object's mass and velocity, and work done on an object can be zero if there is no displacement, as shown in the wall-pushing scenario. Additionally, negative work occurs when a force opposes motion, and the total work done in moving an object up and down can also result in zero. Understanding these principles is crucial for solving related physics problems effectively.
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1) The derivative of work with respect to time is equal to


force

acceleration

momentum

power

impulse

Ans. Power, I think this is correct!

2) If the dot product between two vectors is zero, then the angle between the two vectors is


90°

270°



180°

(a) or (b)

(c) or (d)

Ans. I know it is 90. but can it be 180 also

3) The kinetic energy of an object depends on


its mass.

its velocity.

the direction of its velocity.

it position.

(a) and (b) This is the correct answer.

(a), (b), and (c)

(a) and (d)

4) You push a wall with a force of 11 N for a period of 1 seconds. The wall does not move. The total work done by you is...


11 J.

-11 J.

11 J.

-11 J.

0 J.

I think the answer should be zero , since the wall isn't moving.

5) The units for kinetic energy can be written as...


force/distance.

force * distance.

force * time.

None of the above.

Need help here, also

6) A car is rolling to the right and your try to slow it down by pushing on it to the left. You are doing...


positive work.

negative work.

no work until the car comes to a stop.

You can’t tell because I didn’t tell you which way was positive.

I think it is negative work.

7) You move a box that has a mass of 4 kg in upwards direction 1.1 meters. You then lower it down 1.1 meters to its original position. Assuming that upwards is positive, the total work you did was...


0 J.

4.4 J.

-4.4 J.

8.8 J.

-8.8 J.

None of the above.

I need help here. Can somebody explain too.
8) I can increase the amount of work done on an object by a force by...


increasing the magnitude of the force.

increasing the distance the force moves the object.

increasing the angle between the direction the object moves and the direction of the force.

(a) and (b)

(a) and (c)

All of the above.

None of the above.

Don't have any idea here.
9) Electric motors often give their size in terms of power they produce. If a 110 Watt motor runs for 12 seconds, we can say...


It could have produced a force of 1320 N.

It could have produced a force of 9.16666666666667 N.

It could have done 9.16666666666667 J of work.

It could have done 1320 J of work.

It could have done -9.16666666666667 J of work.

Please help me here.
10) In Touchstone Example 9-1, how much work would have been done by the wind if it had a y component of -7 N. (The motion of the crate remains the same.)


-21 J

21 J

6 J

-6 J

-7 J

7 J
 
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Naeem said:
1) The derivative of work with respect to time is equal to


force

acceleration

momentum

power

impulse

Ans. Power, I think this is correct!

2) If the dot product between two vectors is zero, then the angle between the two vectors is


90°

270°



180°

(a) or (b)

(c) or (d)

Ans. I know it is 90. but can it be 180 also

1. is correct.

2. It's either 90 or 270, as cos 180 = -1 and not 0.
 


I'm not sure about this one, can someone please explain.

Ans. The correct answer is -21 J. The work done by the wind would be equal to the product of the force (-7 N) and the distance the crate moves in the y direction (3 m), which is -21 J. The negative sign indicates that the work is done in the opposite direction of the motion of the crate.
 
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