Question

a large mass m1=5.75 kg and is attached to a smaller mass m2=3.53 kg by a string. The string is hung over a pulley as shown in the figure, and the mass of the pulley and string are negligible compared to the other two masses. Mass 1 is started with an initial downward speed of 2.13 m/s. What is the speed of mass 2 after it has moved ℎ=2.47 meters?

Answer 1

The speed of the second mass after it has moved ℎ=2.47 meters will be 1.09 m/s approximately

What are we to consider in equilibrium ?

Whenever the friction in the pulley is negligible, the two blocks will accelerate at the same magnitude. Also, the tension at both sides will be the same.

Given that a large mass m1=5.75 kg and is attached to a smaller mass m2=3.53 kg by a string and the mass of the pulley and string are negligible compared to the other two masses. Mass 1 is started with an initial downward speed of 2.13 m/s.

The acceleration at which they will both move will be;

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

a = (5.75 - 3.53) / (5.75 + 3.53)

a = 2.22 / 9.28

a = 0.24 m/s²

Let us assume that the second mass starts from rest, and the distance covered is the h = 2.47 m

We can use third equation of motion to calculate the speed of mass 2 after it has moved ℎ=2.47 meters.

v² = u² + 2as

since u =0

v² = 2 × 0.24 × 2.47

v² = 1.1856

v = √1.19

v = 1.0888 m/s

Therefore, the speed of mass 2 after it has moved ℎ=2.47 meters will be 1.09 m/s approximately

Learn more about Equilibrium here: brainly.com/question/517289

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