Four boxes of various masses are placed on a friction
free table. Rank these boxes according to net force
by listing them in order from greatest to least.
A) <-5N--| 5 kg |--10N->
B) <-10N--|10kg|-15N-->
C) <-10N--|5kg|-15N-->
D) <-5N--|20kg|-15N-->
Thank You :)
(the |_ kg | are the boxes)
=)
I apologize I forgot to add: which has the most force to the right?
will this still affect the answer?
Thank You
~Jill :]
If you read my answer (D), it will answer your latest question.
Ok thank you :) I didn't see the rest of the answer because it went off the page, thank you so much for your help! :)
What if box D is 15 kilos?
Well, it seems like these boxes are having quite the forceful party on the table! Let's see who's the strongest in terms of net force:
D) <-5N--|20kg|-15N->: This hefty 20kg box has a force of -5N pulling it to the left and a force of -15N pulling it to the right. So, the net force on this box is (-5N) + (-15N) = -20N.
B) <-10N--|10kg|-15N->: This 10kg box has a force of -10N pulling it to the left and a force of -15N pulling it to the right. The net force on this box is (-10N) + (-15N) = -25N.
C) <-10N--|5kg|-15N->: This 5kg box has a force of -10N pulling it to the left and a force of -15N pulling it to the right. The net force on this box is (-10N) + (-15N) = -25N.
A) <-5N--|5kg|--10N->: This 5kg box has a force of -5N pulling it to the left and a force of -10N pulling it to the right. The net force on this box is (-5N) + (-10N) = -15N.
So, the ranking from greatest to least net force is:
1) D) <-5N--|20kg|-15N-> (-20N)
2) B) <-10N--|10kg|-15N-> (-25N)
3) C) <-10N--|5kg|-15N-> (-25N)
4) A) <-5N--|5kg|--10N-> (-15N)
They certainly know how to make a forceful entrance!
To rank the boxes according to net force, we need to calculate the net force on each box. The net force can be determined by adding up all the forces acting on the box.
Let's calculate the net force for each box:
A) The net force on box A is the difference between the two forces acting on it. The forces are in opposite directions, so we subtract them: 10N - 5N = 5N. The net force on box A is 5N.
B) The net force on box B is the sum of the two forces acting on it. The forces are in the same direction, so we add them: 10N + 15N = 25N. The net force on box B is 25N.
C) The net force on box C is the sum of the two forces acting on it. The forces are in the same direction, so we add them: 10N + 15N = 25N. The net force on box C is also 25N.
D) The net force on box D is the difference between the two forces acting on it. The forces are in opposite directions, so we subtract them: 15N - 5N = 10N. The net force on box D is 10N.
Now that we have calculated the net force for each box, we can rank them from greatest to least:
Ranking: B > C > D > A
So, the correct ranking is Box B with the greatest net force, followed by Boxes C, D, and finally Box A with the least net force.
With no friction, the mass in the box will not affect the net force.
D) will have a net force of 10 N to the right. All of the other three will have a net force of 5 N to the right. They would be "tied"