Question

A 5.0 kg cart on a table is connected to a string that hangs over pulley suspending a 3.0 kg block over the side of the table. Determine the acceleration of the system, as well as the tension in the string. (6 marks)

Answer 1

(a) Applying the Newton's Second Law of Motion we get:

$$\sum F_x=m_{cart}a_x,$$ $$T=m_{cart}a. (1)$$ $$\sum F_y=m_{block}a_y,$$ $$m_{block}g-T=m_{block}a. (2)$$

Let’s substitute $T$ into the equation (2) and find the acceleration of the system:

$$m_{block}g-m_{cart}a=m_{block}a,$$ $$a=\dfrac{m_{block}g}{m_{cart}+m_{block}},$$ $$a=\dfrac{3.0\ kg\cdot9.8\ \dfrac{m}{s^2}}{5.0\ kg+3.0\ kg}=3.675\ \dfrac{m}{s^2}.$$

(b) Finally, we can find the tension in the string:

$$T=m_{cart}a=5.0\ kg\cdot 3.675\ \dfrac{m}{s^2}=18.4\ N.$$