In the ideal Atwood machine arrangement shown, what is the change in kinetic energy (in Joule) of the system in $\frac{1}{2}$ second of their motion starting from rest ?

Question

In the ideal Atwood machine arrangement shown, what is the change in kinetic energy (in Joule) of the system in 2 ​ 1 ​ second of their motion starting from rest ?

TestHub · Accepted Answer

Problem: Atwood Machine - Change in Kinetic Energy

Given:

Masses: $m_{1} = 2$ kg, $m_{2} = 3$ kg

Time: $t = \frac{1}{2}$ seconds

Initial velocity: $v_{0} = 0$ m/s

1. Calculate the net force:

The net force acting on the system is the difference in weight between the two masses:

$F_{n e t} = (m_{2} - m_{1}) g$

$F_{n e t} = (3 k g - 2 k g) \times 10 m / s^{2} = 10 N$

2. Calculate the acceleration:

The acceleration of the system is given by:

$a = \frac{F _{n e t}}{m _{1} + m _{2}} = \frac{10}{2 k g + 3 k g} = 2 m / s^{2}$

3. Calculate the final velocity:

Using the equation of motion:

$v = v_{0} + a t$

Since the system starts from rest, $v_{0} = 0$ :

$v = (2 m / s^{2}) \times \frac{1}{2} s = 2$

4. Calculate the change in kinetic energy:

$Δ K E = \frac{1}{2} \times 5 \times (2)^{2} = \frac{1}{2} \times 5 \times 2 = 5 J$

FINAL ANSWER: $Δ K E = 5$ J

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