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Two students are pushing identical crates across a frictionless floor. The crates are initially at rest. Bing applies a horizontal force of 23 N to his crate. Bob, who is shorter than Bing, pushes on his crate at an angle of 33° below the horizontal, also with a force of 23 N. Find the ratio of the masses of the two crates if Bing’s crate is moving at twice the speed of Bob’s crate after they have traveled a distance of 33.5 m across the floor? (20 pts)

Respuesta :

Answer:

[tex]\frac{m_1}{m_2} = 0.298[/tex]

Explanation:

As we know that both the crates are moving at different speed after moving 33.5 m

speed of bling crate = 2 (speed of bob crate)

now we know that

Bling apply force horizontal so the acceleration of his crate is given as

[tex]a_1 = \frac{23}{m_1}[/tex]

now force of Bob is at 33 degree with the horizontal

so the acceleration of crate of Bob is given as

[tex]a_2 = \frac{23 cos33}{m_2}[/tex]

now we know that work done by Bling and bob on their crate is equal to the kinetic energy of the crate

so for Bling's and Bob'scrate we can say

[tex]F. d = \frac{1}{2}m v^2[/tex]

now the ratio of the work done of two is given as

[tex]\frac{W_1}{W_2} = \frac{0.5 m_1 v_1^2}{0.5 m_2 v_2^2}[/tex]

[tex]\frac{23 d}{23 d cos33} = \frac{m_1 (2v)^2}{m_2(v^2)}[/tex]

[tex]\frac{1}{cos33} = \frac{4 m_1}{m_2}[/tex]

[tex]\frac{m_1}{m_2} = 0.298[/tex]