Question

g Question 11 pts Consider two masses connected by a string hanging over a pulley. The pulley is a uniform cylinder of mass 3.0 kg. Initially m1 is on the ground and m2 rests 2.9 m above the ground. After the system is released, what is the speed of m2 just before it hits the ground? m1= 30 kg and m2= 35 kg Group of answer choices 2.1 m/s 1.4 m/s 9.8 m/s 4.3 m/s 1.9 m/s

Answer 1

Answer:

The speed of m2 just before it hits the ground is 2.1 m/s

Explanation:

mass on the ground m1 = 30 kg

mass oat rest at the above the ground m2 = 35 kg

height of m2 above the ground =2.9 m

Let the tension on the string be taken as T

for the mass m2 to reach the ground, its force equation is given as

$m_{2} g - T = m_{2}a$    ....equ 1

where g is acceleration due to gravity = 9.81 m/s^2

and a is the acceleration with which it moves down

For mass m1 to move up, its force equation is

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

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

$T = m_{1}(a + g)$    ....equ 2

substituting T in equ 1, we have

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

imputing values, we have

 $(35*9.81) - 30(a+9.81) = 35a$

 $343.35 - 30a-294.3 = 35a$

$343.35 -294.3 = 35a+ 30a$

$49.05 = 65a$

a = 49.05/65 = 0.755 m/s^2

The initial velocity of mass m2 = u = 0

acceleration of mass m2 = a = 0.755 m/s^2

distance to the ground = d = 2.9 m

final velocity = v = ?

using Newton's equation of motion

$v^{2}= u^{2} + 2ad$

substituting values, we have

$v^{2}= 0^{2} + 2*0.755*2.9$

$v^{2}= 2*0.755*2.9 = 4.379\\v = \sqrt{4.379}$

v = 2.1 m/s