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a 5 kg box slides down a 3 meter long ramp angled at 30 degrees. the box feels a constant 10 n force of friction. if the box starts at rest, how fast is it going at the bottom of the ramp?

Respuesta :

The box is going at the bottom of the ramp at the speed of 4.24m/s

given:

The mass of the box = 5kg

The incident angle of the ramp = 30degrees

Friction force = 10 newton

The formula is

              net force = mgsinθ - friction force

                  ma = mgsinθ - friction force

                   a = (5x5 - 10)5

               acceleration a = 3 m/s^2

The equation of the motion

                    v^2 = u^2 + 2as

                   v^2 =  0 = 2x3x3

                  v  =  4.24 m/s  

To know more about Friction force, click below link

https://brainly.com/question/1714663?referrer=searchResults

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