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A 39.1 kg woman wearing ice skates is standing on a skating rink next to a vertical wall. She pushes off against the wall and begins to slide along the perfectly horizontal surface of the rink. The coefficient of kinetic friction between the blades of her ice skates and the surface of the rink is 0.04. If she slides a distance of 14.8 m before stopping and her push against the wall lasted 0.2 s, what was the average force (in N) that she exerted on the wall?

Respuesta :

chamberlainuket

Answer:

666.67N

Explanation:

from work energy principle

work done by friction = change in KE

-uk*m*g*x = (1/2)*m*(vf^2-vi^2)

-uk*g*x = (1/2)*(vf^2-vi^2)

as the woman is stopped the final velocity vf = 0

-0.04*9.8*14.8 = (1/2)*(0-vi^2)

initial speed vi = 3.41 m/s

after pushing against the wall

momentum gained by woman = m*vi

force applied F = m*vi/t

F = 39.1*3.41/0.2

F = 666.66 N <<

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