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A 0.095-kg dodge ball is traveling at 22 m/s along the concrete ground. Friction
with a magnitude of 0.75 N is exerted on the moving ball by the surface of the
concrete. Someone also uses their foot to slow the motion of the ball to a stop. If
the ball travels 0.20 m before stopping, then determine the amount of force
applied to the ball by the foot.

Respuesta :

indrawboy

vf² = vi² - 2ad ( a= -, negative sign = deceleration)

vf = 0 (stop)

vi² = 2ad

22² = 2 x a x 0.2

a = 1210 m/s²

∑F= m.a

F - 0.75 = 0.095 x 1210

F = 115.7 N

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