Question

Two masses of 2kg and 4kg are connected by a light string passing over a light frictionless pulley. If the system is released from rest determine the velocity of the 4kg mass when it has descended a distance of 1.4m

Answer 1

Answer on Question#41667, Physics, Mechanics

Two masses of $2\mathrm{kg}$ and $4\mathrm{kg}$ are connected by a light string passing over a light frictionless pulley. If the system is released from rest determine the velocity of the $4\mathrm{kg}$ mass when it has descended a distance of $1.4\mathrm{m}$ .

Solution:

Given:

$m_{1} = 4\mathrm{kg}$

$m_{2} = 2\mathrm{kg}$

$W_{1} = m_{1}g$ (weight of first body)

$W_{1} = m_{2}g$ (weight of second body)

$h = 1.4\mathrm{m}$

$v_{1} = ?$

The equations of motion are:

$m_{1}a = m_{1}g - T$

$m_{2}a = T - m_{2}g$

where $T$ is a tension of the string, and its remains the same at any point.

The adding of two equations gives:

$m_{1}a + m_{2}a = m_{1}g - T + T - m_{2}g$

$m_{1}a + m_{2}a = m_{1}g - m_{2}g = g(m_{1} - m_{2})$

Thus, the acceleration is

$$a = \frac {g (m _ {1} - m _ {2})}{m _ {1} + m _ {2}}$$ $$a = \frac {9.81 \cdot (4 - 2)}{4 + 2} = 3.27 \mathrm{m/s^2}$$

The kinematic equation that describes an object's motion is:

$$v _ {1} ^ {2} = v _ {0} ^ {2} + 2 a h$$

The symbol $h$ stands for the displacement of the object. The symbol $a$ stands for the acceleration of the object. The initial velocity $v_{0} = 0$ .

Thus, $v_{1} = \sqrt{2ah} = \sqrt{2 \cdot 3.27 \cdot 1.4} = 3.03 \approx 3 \, \text{m/s}$

Answer. $v_{1} = 3 \, \text{m/s}$ .

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