I've got a bunch of questions, that I appear to have gotten wrong previously..

1. If you use a horozontal force of 30.0 N to slide a wooden crate of 12.0 kg across a floor at a constant velocity, what is the coefficient of sliding friction between crate and floor?

2. The instruments attached to a weather balloon have a mass of 5.0 kg.

a)the balloon is released and exerts an upward force of 98N on the intruments. What is the acceleration of the balloon and instruments?

b)after the balloon has accelerated for 10s, the instruments are released. What is the velocity of the instruments at the moment of their release?

c)what net force acts on the instruments after their release?

d)when does the direction of their velocity first become downward?

3. A 0.25 kg soccer ball is rolling at 6.0 m/s toward a player. The player kicks the ball back in the opposite direction and gives it a velocity of -14 m/s. What is the average force during the interaction between the player's foot and the ball if the interaction lasts 2.0 X 10-2 s.

4. A person weighing 490N stands on a scalein an elevator.
a) what does the scale read when the elevator is at rest?
what is the reading on the scale when the elevator rises at a constant velocity?
c) the elevator slows down at -2.2 m/s2 as it reaches the desired floor. What does the scale read?

5. A 50-m wide river flows east at 5.0 m/s. A boat heads south across the river at 7.8 m/s.
a) calculate the magnitude and direction of the resultant velocity.
calculate where the boat will dock on the other side of the river.

6. A sign that weighs 648 N is suspended by two ropes that make an angle of 38 degrees with the horozontal. What force does each rope exert on the sign?

7. A 46N force acts S32 degrees W while a 56N force acts S40 degrees E. Calculate the magnitude and direction of the equilibrant.

I am not going to answer these questions but will try and get you started.

In the first question the key is that the force produces no acceleration, in other words the frictional force is equal to the pushing force.

2nd Q - force up = 98N, force down = 5.0*g = 48N hence net upward force producing an acceleration.

3rd Q - Impulse equation is F*t = change in momentum.

4th Q - when the lift is accelerating there is a extra force on the mass on the scale - when it moves at constant velocity there isn't.

5th Q, and 7th seem fairly straightforward vector problems. Can you visualise the boat being carried downstream? Draw a picture and use trigonometry.

6th Q - you need a force diagram showing the mass acting downwards and the tension in the cables acting upwards, but at an angle. Again a diagram should illustrate that you need the cosine of the angle to determine the vertical force.


Someone may be in to answer your problems in more detail but as you haven't shown any thoughts or even told us what you are having trouble with I think I have spent enough time typing.
In fact I suggest that if you read the thread at the top of this board then you will find some of the answers yourself.

If you got them wrong before, why not post a few notes on what you did the first time. Otherwise we're all going to think you just want us to do your homework for you!