Question

Naomi is moving two boxes (A = 20kg and B = 10kg), one on top of the other, with
magnitude F = 150 N along the horizontal. The coefficient of kinetic friction between box A
and the floor is 0.40 while the coefficient of static friction between box A and box B is 0.80. What is the acceleration of the boxes.
a. 1.1 m/s2
c. 5.0 m/s2
b. 8.9 m/s2
d. 2.9 m/s2

Answer 1

please help me @ hoslos

Answer 2

@helpme25 @hoslos

Answer 3

First of all, to get the exact acceleration n of the boxes, the force applied has to be the one that will not make them slide over each other. To do that, we need to work with the upper box(B):
Normal contact force or N= mg or 10*9.81=98.1N
Ff(frictional force) = required applied force to avoid sliding=€N or 0.8*98.1=78.48N
Do you have the diagram?

Answer 4

nope.

Answer 5

i have sorry

Answer 6

Hehe, ok. Please post it out.

Answer 7

Do they mention anything about sliding or no sliding of boxes?

Answer 8

nope :D

Answer 9

Guess not. ALRIGHt forget about what I wrote ealier. Let us go for the case of the boxes and floor. First we find the frictional force, find N = mg or (10+20)*9.81=294.3N
The frictional force is then $N or 0.4*294.3=117.72N.
Use the 2: law of motion's formulaof F-R=ma or 150-117.72=(20+10)a and a is 1.076, approximately 1.1m/s^2

Answer 10

where did u get the R?

Answer 11

R is the resistive force or frictional force that I calculated right there.