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 :)
I forgot to include:
Which one has the most force to the right?
I appreciate it! :]
To determine the net force acting on each box, we need to add up the forces acting on them. Net force is given by the equation: net force = sum of all forces.
Let's go through each box one by one:
A) The forces acting on box A are -5N to the left and 10N to the right. To calculate the net force, we subtract the force to the left from the force to the right: net force of box A = 10N - 5N = 5N.
B) The forces acting on box B are -10N to the left and 15N to the right. The net force of box B is calculated the same way: net force of box B = 15N - 10N = 5N.
C) For box C, the forces are -10N to the left and 15N to the right. The net force of box C is: net force of box C = 15N - 10N = 5N.
D) The forces acting on box D are -5N to the left and 15N to the right. Calculate the net force of box D: net force of box D = 15N - 5N = 10N.
So, in terms of the net force from greatest to least: D > A = B = C.
To answer the additional question, "Which one has the most force to the right?", we can see that box D has the highest net force to the right with 10N.