contestada

A mass m = 0.3 kg is released from rest at the origin point 0. The mass falls under the influence of gravity. When the mass reaches point A, it has a velocity of v downward and when the mass reaches point B its velocity is 5v. What is the distance between points A & B divided by the distance between points 0 & A?

Respuesta :

Answer:

24

Explanation:

The mass = 3 kg

at point O all the mechanical energy of the system is due to its potential energy PE. The body is at rest.

PE = mgh

but ME = PE + KE = constant   (law of energy conservation)

KE is the kinetic energy

since KE is zero at this point, then,

ME = mgh

where m is the mass

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

h is the height = O

ME = 3 x 9.81 x O

ME = 29.43-O

At point A the total ME is due to its PE and its kE

PE at this point = mgh = 3 x 9.81 x A = 29.43-A

KE = [tex]\frac{1}{2}mv^{2}[/tex]

velocity = v at this point, therefore,

KE = [tex]\frac{1}{2}*3*v^{2}[/tex] = [tex]\frac{3}{2} }v^{2}[/tex]

therefore,

ME = 29.43-A + [tex]\frac{3}{2} }v^{2}[/tex]

Equating ME for the points O and A, we have

29.43-O = 29.43-A + [tex]\frac{3}{2} }v^{2}[/tex]

29.43-O - 29.43-A = [tex]\frac{3}{2} }v^{2}[/tex]

(O - A)29.43 = [tex]\frac{3}{2} }v^{2}[/tex]

O - A = 0.051[tex]v^{2}[/tex]   this is the distance between point O and A

For point B

PE = 29.43-B

KE = [tex]\frac{25}{2}*3*v^{2}[/tex] = 37.5[tex]v^{2}[/tex]        (velocity is equal to [tex]5v[/tex] at this point)

therefore,

ME = 29.43-B + 37.5[tex]v^{2}[/tex]

Equating the ME for points A and B, we have

29.43-A + [tex]\frac{3}{2} }v^{2}[/tex] = 29.43-B + 37.5[tex]v^{2}[/tex]

29.43-A - 29.43-B = 37.5[tex]v^{2}[/tex] - [tex]\frac{3}{2} }v^{2}[/tex]

(A - B)29.43 = 36[tex]v^{2}[/tex]

A - B = 1.22[tex]v^{2}[/tex]    this is the distance between points A and B

The distance between points A & B divided by the distance between points 0 & A will be

1.22[tex]v^{2}[/tex]/0.051 = 23.9 ≅ 24