- #1
Zhalfirin88
- 137
- 0
Homework Statement
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.
The Attempt at a Solution
So I did that, but ended up wrong.
[tex] \Sigma F = ma [/tex]
[tex] -f_k + mg = ma [/tex] Because the tension forces would cancel.
[tex] \frac{-(1.1kg * 9.8\frac{m}{s^2}) + (3.20kg * 9.8\frac{m}{s^2})}{(1.1kg + 3.2kg)} [/tex]
[tex] a = 4.786 \frac{m}{s^2} [/tex]
[tex] v_f^2 = v_o^2 + 2a\Delta x [/tex]
[tex] v_f = \sqrt{2 * 4.786\frac{m}{s^2} * .47m} [/tex]
[tex] v_f = 2.12 \frac{m}{s} [/tex]
But this was wrong, so where did I go wrong?