How much sooner does the box reach the bottom of the incline than the disk? express your answer in terms of some or all of the variables m, h, θ, and r, as well as the acceleration due to gravity g. view available hint (s) ?

Question

Heidi Vega

Physics

13 November, 22:45

How much sooner does the box reach the bottom of the incline than the disk? express your answer in terms of some or all of the variables m, h, θ, and r, as well as the acceleration due to gravity g. view available hint (s) ?

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Answers (1)

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George Jacobs · Answer 1

The expression for time difference is

Δt = td - tb

Substitute 1/sinФ√3h/g for td and 1/sinФ√2h/g for tb

Δt = 1/sinФ√3h/g - 1/sinФ√2h/g

=1/sinФ√h/g (√3 - √2)

= 0.32 [1/sinФ √h/g]

The time where by the lock comes down to the bottom of inclined before the disk is 0.32 [ 1 / sinФ √h/g]