# Forces and Tension

## Homework Statement

The 1.0 kg block in the figure is tied to the wall with a rope. It sits on top of the 2.0 kg block. The lower block is pulled to the right with a tension force of 20 N. The coefficient of kinetic friction at both the lower and upper surfaces of the 2.0 kg block is mu_k = 0.380.

What is the acceleration of the 2.0 kg block?

## Homework Equations

a = F/m
F_friction = mg * mu_k

## The Attempt at a Solution

I found in a previous question that the tension of the rope holding the 1kg block, and thus the force of friction on the top of the 2kg block, was 3.72N (mg * mu_k = 9.8 * .38 = 3.72).

So the frictional force on the bottom of the 2kg block:

mg * mu_k = 2 * 9.8 * .38 = 7.448

Those two forces hold back the 20N force pulling to the right, so the total force acting on it to the right is:

20 - 3.72 - 7.448 = 8.832

So using a = F/m:

a = 8.832 / 2 = 4.416 m/s^2

However, 4.416 is incorrect. Any help you could give would be appreciated. Thanks!

Last edited:

Related Introductory Physics Homework Help News on Phys.org
The upper friction done by the 1-kg block is okay (3.72N), but I see a problem in the friction on the bottom, you should check what's the real mass affecting to that second friction.

rl.bhat
Homework Helper
Hi Dormin, welcome to PF.
Net friction force on the lower surface is due to the normal reaction of 3 kg mass. It is in the opposite direction of 20 N. Frictional force on the upper surface of 2 kg block is due to 1 kg mass. It is also in the opposite direction of applied force.
Now find the net friction force on the lower block which opposes the applied force.