Question

A 3.00-kg mass rests on the ground. It is attached to a string which goes vertically to and over an ideal pulley. A second mass is attached to the other end of the string and released. The 3.00-kg mass rises 50.0 cm in 1.00 s. How large was the other mass?
A) 3.67 kg
B) 4.29 kg
C) 6.83 kg
D) 7.15 kg
E) 7.34 kg

Answer 1

Answer:A

Explanation:

mass of object=3 kg

distance moved=50 cm

time t=1 s

$s=ut+\frac{at^2}{2}$

$0.5=0+\frac{a\cdot 1^2}{2}$

$a=1 m/s^2$

Let T is the tension in rope

$T-mg=ma$

Let M be the other mass

$Mg-T=Ma$

$T=M(g-a)$

$M=m\frac{g+a}{g-a}$

$M=3\times \frac{9.8+1}{9.8-1}$

$M=3\times 1.227$

$M=3.67 kg$

Answer 2

The size of the mass is mathematically given as

m2=3.68kg

Option A is correct

What is size of the mass?

Question Parameter(s):

A 3.00-kg mass rests on the ground.

The 3.00-kg mass rises 50.0 cm in 1.00 s.

Generally, the equation for the Motion  is mathematically given as

$s=ut+\frac{at^2}{2}$

Therefore

$0.50=0+\frac{a\cdot 1^2}{2}$

a=1m/s^2

T-m1g=m1a

Therefore, Tension

$m_2=m_1\times \frac{g+a}{g-a}$

$m_2=3\times \frac{10.8}{8.8}$

m2=3.68kg

Read more about mass

https://brainly.com/question/15959704