I have a question about Newton's law.
The question says
block A(mass 2.25kg) rests on a tabletop. It is connected by a horizontal cord passing over a light, friction less pulley to hanging block B(mass 1.30kg). The coefficient of kinetic friction between block A and the tabletop is 0.450. After the blocks are released from rest, find (a) the speed of each block after moving 3.00 cm and (b) the tension in the cord. Include the free-body diagram or diagrams you used to determine the answer.
I tried to solve using following formula
$$f_{k} = \mu_{k}N$$
$$T-w_{B} = m_{b}a$$
$$T = m_{A}a$$
First time, I tried to get $a$ Since
$$f_{k}=\mu_{k}N$$
$$m_{A}a = \mu_{k}N$$
$$m_{A}a = \mu_{k}m_{A}g$$
From given $m_{A} = 0.45, \delta x=0.03(m), m_{A}=2.25$
I cancelled out $m_{A}$ both side then I got
$$a = \mu_{k}g$$ which is $$a = 0.45*9.8 \implies a = 4.41\ \mathrm{m/s^2}$$
Part a asked for its speed so I used $\delta x = \frac{V^2-V_{0}^2}{2a}$
I got $V = 0.897\text{ m/s}$
but the speed I got was not right speed. So I think I made some mistake at some point. What did I do wrong?
Best Answer
Your $ T=m_Aa $ is wrong.
Because Tension due to $ m_B $ is pulling it forward and friction is trying to resist that force.
Using $v^2 = u^2 + 2as$ for calculating velocity after moving $ s = 0.03 m $ forward with initial velocity $ u = 0 m/s $ is correct!.