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A 1 kg object starts 3 meters from the ground on a 24 degree incline with a friction coefficient of 0.12. When it hits the grou...Asked by matt
A 1 kg object starts 3 meters from the ground on a 24 degree incline with a friction coefficient of 0.12. When it hits the ground, it collides and sticks to a 8 kg object. The coefficient friction on the horizontal surface is 0.24 How far do the objects travel away from the incline?
When calculating the frictional work along the incline plane would this be the formula W= -0.12(1)(9.8)(3cos24)
When calculating the frictional work along the incline plane would this be the formula W= -0.12(1)(9.8)(3cos24)
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Answered by
drwls
You got the distance travelled wrong and the friction force wrong.
In traveling from 3 meters ABOVE the ground on a 24 degree incline, the distance travelled is 3/sin 24 = 7.38 meters. The opposing friction force is
M g cos24*0.12 = 1.07 N. Available kinetic energy at the bottom is
M g*3.0 m - 1.07N*7.38 m = 21.5 J
Check my numbers
In traveling from 3 meters ABOVE the ground on a 24 degree incline, the distance travelled is 3/sin 24 = 7.38 meters. The opposing friction force is
M g cos24*0.12 = 1.07 N. Available kinetic energy at the bottom is
M g*3.0 m - 1.07N*7.38 m = 21.5 J
Check my numbers
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