Having trouble with equations pertaining to Newton's Laws?

Question

Hello, all. I need a little guidance on solving a Physics problem. I don't want anyone to do it for me or give me the answers -- I need some help, though in understanding how to solve it. Please let me know what equations I need in order to find the answers to the following:

A rope exerts a constant horizontal force of 250 N to pull a 60-kg crate across the floor. The velocity of the crate is observed to increase from 1 m/s to 3 m/s in a time of 2 seconds under the influence of this force and the frictional force exerted by the floor on the crate.

a.) What is the acceleration of the crate?
b.) What is the total force acting upon the crate?
c.) What is the magnitude and frictional force acting on the crate?
d.) What force would have to be applied to the crate by the rope in order for the crate to move with constant velocity? Explain.

Thanks in advance for any help on this!

I was trying to figure out acceleration by a = F/m. for Part B, should I add the force of gravity when summing up all forces acting on the crate? If so, how do I determine what the frictional force is - it can't be the same as gravity, can it? Because the crate DID move.

Okay, I probably will need some clarification. I actually started out understanding pretty much what everone has said, here, but there a few elements that does not make sense to me. Firstly, since the book stated the velocity as going from 1 to 3 m/s, why is the result 1 m/s squared and not just 1m/s? Next, I know that F=ma, which means that here F=60kg(1 m/s) or 60N, but then we have the 250N rope pull. So, wouldn't the total Force be 310N? Furthermore, I don't "get" why the frictional force would be -250N. If it cancels out the rope pulling, it would mean that the crate didn't move. The cart did move, but because of the rope -- not its own weight. I would think teh frictional force is LESS than -250N, because the crate moved. And what is its magnitude? Furthermore, the last question (d.) is confusing because the problem says the crate did move at a constant rate.

Answer 1

a) The acceleration is the change in velocity divided by the time over which this change takes place. a = (3 m/s - 1 m/s) / 2 s.
b) The total force acting on the crate is the mass of the crate (60 kg) times the acceleration of the crate, which you just calculated in a) above. F = m * a
c) You exert a force of 250 N on the crate, but the total force on the crate is what you just calculated in b) above. There's two forces on the crate, the 250 N force from the rope and the force due to friction. The sum of these two forces, which act in opposite directions, equal the total force you calculated in b). F = F(rope) - F(friction); F(friction) = F - F(rope) = F - 250N.
d) The crate will move with constant velocity, i.e no acceleration, when the force applied by the rope exactly equals the force due to friction, which you just calculated in c) above. F(no acceleration) = F(friction).

Additional details
I was trying to figure out acceleration by a = F/m. for Part B, should I add the force of gravity when summing up all forces acting on the crate?
If so, how do I determine what the frictional force is - it can't be the same as gravity, can it? Because the crate DID move.
No. In this instance, the force of gravity doesn't come into it. Remember that the crate is moving horizontally, but the force of gravity acts vertically - there's no horizontal component of gravity. While it's true that the force of gravity is acting on the crate, there's a reactive force of the ground pushing up on the crate that exactly counters it, so the crate doesn't move up or down. In this problem, the only forces to consider are the unbalanced ones, which are the tension in the rope and the force due to friction.

Firstly, since the book stated the velocity as going from 1 to 3 m/s, why is the result 1 m/s squared and not just 1m/s?
Be careful how you write it - here's where not having access to math notation to answer these questions can get in the way. You have a change in velocity from 1 m/s to 3 m/s, which occurs in 2 seconds. Acceleration is the change in velocity divided by the time in which the change occurs, so you get (3 m/s-1 m/s)/2s, or (2 m/s)/2s, which is 1 m/(seconds squared). Make sure you're not confusing that with (1 m/s) squared. The squared sign just goes with the seconds part, so you get units of acceleration.

Next, I know that F=ma, which means that here F=60kg(1 m/s) or 60N, but then we have the 250N rope pull. So, wouldn't the total Force be 310N?
No. Remember that forces are vectors, and the total force is what causes the acceleration. You're correct in the first part; F = 60 kg * 1 m/(s^2) = 60 N. That's the result of a 250 N force acting in this direction ------------> combined with a 190 N force acing in this direction <---------, which is coming from friction. Add the vectors together, and you get a total of 60 N, which is what causes the acceleration. Note that there's still gravity and the reactive force of the floor countering it, but they still don't have any part in the sideways motiion.

Furthermore, I don't "get" why the frictional force would be -250N. If it cancels out the rope pulling, it would mean that the crate didn't move. The cart did move, but because of the rope -- not its own weight. I would think teh frictional force is LESS than -250N, because the crate moved. And what is its magnitude?
You're correct here; the frictional force is not -250 N, it's -190 N. You can tell that because the total force is 60 N. You're right in saying that if the frictional force was -250 N, the crate would not move.


Furthermore, the last question (d.) is confusing because the problem says the crate did move at a constant rate.
No, it doesn't; it says the rate changes from 1 m/s to 3 m/s in 2 s. In order for the crate to not undergo acceleration, the total force would have to be 0. You know the frictional force is -190 N, so if the force from the rope was +190 N, the crate would move at constant velocity but would not accelerate.

Drop me an e-mail if anything's still not clear.

Answer 2

A rope exerts a constant horizontal force of 250 N to pull a 60-kg crate across the floor. The velocity of the crate is observed to increase from 1 m/s to 3 m/s in a time of 2 seconds under the influence of this force and the frictional force between the floor and the crate. What is the magnitude of the frictional force acting on the crate? Use g = 10 m/s/s.

Answer 3

To answer your additional questions:
Gravity acts perpendicular to the other forces and has no effect on the motion. You need to figure the acceleration from the change in velocity over time - a = Δv/Δt. Then, knowing the mass and the acceleration, you can calculate the force. The friction force is the total force minus the 250N from the rope. It will be negative because it acts in the opposite direction.

So, sum of forces in the x direction is rope pull minus friction; in the y direction, gravity and the reaction from the floor sum to zero; and in the z direction, nothing.

Answer 4

Acceleration is change of speed per unit of time
What was the change in the speed in 2 seconds?
Then what was the change in 1 second?

Force = mass x acceleration
The mass is 60 kg, the acceleration you just figured.

Now you know the total force on the crate.
The rope force is 250 N, the friction force goes the other way.
Force = 250 - friction force

Answer 5

a)Acceleration is change in velocity over time taken

(3-1)/2 = 1m/s^2

b) This has 2 methods

1) F=ma
F=60x1
F=60

2)total force= friction plus driving.

From part c, friction is 190N, so 250 +-190 =60N
c)As F=ma, resultant force=mass x acceleration

Driving force-friction= 60 x 1

250-z=60
z=250-60
z=190 N against motion, or -190N

d) Constant velocity = no resultant force = friction is same as driving

Therefore friction must equal 250N in the opposite direction, which can be said as -250N.

Hope this helps

Answer 6

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