# Use Newton's second law to find the net force acting on both blocks

1. Sep 23, 2009

### Zhalfirin88

1. The problem statement, all variables and given/known data
Two blocks with masses m1 = 1.10 kg and m2 = 3.20 kg are connected by a massless string. They are released from rest. The coefficient of kinetic friction between the upper block and the surface is 0.490. Assume that the pulley has a negligible mass and is frictionless, and calculate the speed of the blocks after they have moved a distance 47.0 cm.

Hint: Use Newton's second law to find the net force acting on both blocks. Since they are connected by a string, they act as one body and have the same acceleration.

3. The attempt at a solution

So I did that, but ended up wrong.

$$\Sigma F = ma$$

$$-f_k + mg = ma$$ Because the tension forces would cancel.

$$\frac{-(1.1kg * 9.8\frac{m}{s^2}) + (3.20kg * 9.8\frac{m}{s^2})}{(1.1kg + 3.2kg)}$$

$$a = 4.786 \frac{m}{s^2}$$

$$v_f^2 = v_o^2 + 2a\Delta x$$

$$v_f = \sqrt{2 * 4.786\frac{m}{s^2} * .47m}$$

$$v_f = 2.12 \frac{m}{s}$$

But this was wrong, so where did I go wrong?

2. Sep 23, 2009

### Furby

Re: Pulley

It looks like you had friction in your equation (-fk) but did nothing with it. I think you calculated the velocity of the blocks without any friction.

To be honest, I don't really understand the problem. What does it mean by 'between the upper block and the surface'?

3. Sep 23, 2009

### Zhalfirin88

Re: Pulley

Well we had a picture, and I don't know how to transfer it over. Basically m1 is on a surface with a string attached to m2. m2 isn't on a surface. The only forces acting on m2 are mg pulling it down and the tension pulling it up.

Think of a cliff, where m1 is on the cliff and m2 is not on the cliff.

Edit: Aha! found a picture

http://spiff.rit.edu/classes/phys311/workshops/w5b/level_atwood/cart_and_weight.gif

4. Sep 23, 2009

### Furby

Re: Pulley

Yeah, that was my assumption. = ), just wanted to be clear, especially that the upper block=m1.

Try finding the force of friction applied to m1.

Fk=uk*FN

This is the force applied opposite to the direction of m1's motion.

5. Sep 23, 2009

### Zhalfirin88

Re: Pulley

$$\frac{-(1.1kg * 9.8\frac{m}{s^2} * .49) + (3.20kg * 9.8\frac{m}{s^2})}{(1.1kg + 3.2kg)}$$

$$a = 6.065 \frac{m}{s^2}$$

$$v_f^2 = v_o^2 + 2a\Delta x$$

$$v_f = \sqrt{2 * 6.065\frac{m}{s^2} * .47m}$$

$$v_f = 2.388 \frac{m}{s}$$ And that was correct! Just forgot to multiply by the coefficient of friction, thanks!