urgent help please?

Question

two blocks, one 8.0 kg and the other 5.0 kg, sit side by side in contact with each other on a frictionless horizontal surface. if a constant horizontal force of 40N is applied to the 8-kg block in the direction of the smaller block, what is the magnitude of the resulting acceleration? find the contact force the two blocks exert on each other.

Answer 1

F=MA

Your force is 40 N.

Mass is the total of the two blocks, 13 kg.

When you solve for acceleration, use that figure to find the contact force.

Answer 2

Ok, let's solve it :

Let's first work with the first block of 8 kg :

There is the force : F = 40 N

So, applying Newton's second law :

40 - Fc = 8*a

* Note : Fc = compress force

a = acceleration

For the second block : m2 = 5 kg

Applying Newton's second law :

Fc = 5*a

And : 40 - Fc = 8a

then : 13a = 40

a = 3.076 m/s^2

That's the acceleration of the system

The compress force : Fc = 5*3.076 = 15.38 Newtons

Hope that helps

Answer 3

F=m*a
Since the blocks are in contact, the acceleration of both blocks will be equal , so
(8+5)*a=40
a=40/13

the force between the blocks will be related to the acceleration of the small block as
5*a=F
5*40/13=F

j

Answer 4

acceleration = total force / total mass

= 40/13 N

To get the contact force, you can just multiply that acceleration by the mass of the second block (because the contact force is the ONLY force on that block causing its acceleration).