contestada

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?

Respuesta :

Answer:3.68 kg

Explanation:

Given

mass [tex]m_1=3 kg[/tex]

distance traveled by [tex]m_1[/tex] is 50 cm in 1 s

using [tex]s=ut+\frac{at^2}{2}[/tex]

[tex]0.50=0+\frac{a\cdot 1^2}{2}[/tex]

[tex]a=1 m/s^2[/tex]

Suppose T is the tension in string and [tex]m_2[/tex] is the mass of other body

For [tex]m_1[/tex]

[tex]T-m_1g=m_1a[/tex]

[tex]T=m_1(g+a)[/tex]

for other body

[tex]m_2g-T=m_2a[/tex]

[tex]T=m_2(g-a)[/tex]

Equating value of Tension

[tex]m_2=m_1\times \frac{g+a}{g-a}[/tex]

[tex]m_2=3\times \frac{10.8}{8.8}[/tex]

[tex]m_2=3.68 kg[/tex]

Otras preguntas