In addition: To the nearest newton in problem 5 what force does the blue box exert on the red box?
The first answer is a = 8.38 m/s^2
290 = (20.1 + 14.5)a
Solve for a. Can't seem to figure out how to find the force on the red box.
The red box has a mass of 20.1 kg and the blue box has a mass of 14.5 kg and the force is 290 N. To the nearest tenth of a m/s2 what is the acceleration of the combination?
The diagram looks like this:
F --> [Red Box][Blue Box]
On a straight plane, not on an incline.
2 answers
Figured it out.
The answer is N = -122.
F = ma
In this case, acceleration is negative because it's in the opposite direction (being pushed to the right, but we want to know the effects on the red box which is to the left), therefore m • a = 14.5 • -8.38 = -121.5 N.
The answer is N = -122.
F = ma
In this case, acceleration is negative because it's in the opposite direction (being pushed to the right, but we want to know the effects on the red box which is to the left), therefore m • a = 14.5 • -8.38 = -121.5 N.