Question

Two masses of 3 kg and 5 kg are connected by a light string that passes over a smooth polley as shown in the Figure.

Answer 1

i. $T = 36.8\:\text{N}$

ii. $a = 2.45\:\text{m/s}^2$

iii. $x = 1.23\:\text{m}$

Explanation:

Let's write Newton's 2nd law for each object. We will use the sign convention assigned for each as indicated in the figure. Let T be the tension on the string and assume that the string is inextensible so that the two tensions on the strings are equal. Also, let a be the acceleration of the two masses. And $m_1 = 3\:\text{kg}$ and $m_2 = 5\:\text{kg}$

Forces acting on m1:

$T - m_1g = m_1a\:\:\:\:\:\:\:(1)$

Forces acting on m2:

$m_2g - T = m_2a\:\:\:\:\:\:\:(2)$

Combining Eqn(1) and Eqn(2) together, the tensions will cancel out, giving us

$m_2g - m_1g = m_2a + m_1a$

or

$(m_2 - m_1)g = (m2 + m_1)a$

Solving for a,

$a = \left(\dfrac{m_2 - m_1}{m_2 + m_1}\right)g$

$\:\:\:\:= \left(\dfrac{5\:\text{kg} - 3\:\text{kg}}{5\:\text{kg} + 3\:\text{kg}}\right)(9.8\:\text{m/s}^2)$

$\:\:\:\:= 2.45\:\text{m/s}^2$

We can solve for the tension by using this value of acceleration on either Eqn(1) or Eqn(2). Let's use Eqn(1).

$T - (3\:\text{kg})(9.8\:\text{m/s}^2) = (3\:\text{kg})(2.45\:\text{m/s}^2)$

$T = (3\:\text{kg})(9.8\:\text{m/s}^2) + (3\:\text{kg})(2.45\:\text{m/s}^2)$

$\:\:\:\:= 29.4\:\text{m/s}^2 + 7.35\:\text{m/s}^2 = 36.8\:\text{N}$

Assuming that the two objects start from rest, the distance that they travel after one second is given by

$x = \frac{1}{2}at^2 = \frac{1}{2}(2.45\:\text{m/s}^2)(1\:\text{s})^2 = 1.23\:\text{m}$