a body is attached to alower end of aspring suspended from the foot of a lift .when the lift is at rest,a body produces an elongation of 2cm in the spring. when the lift moves downward with an acceleration ,the elongation is changed to 1.95 cm. find the acceleration of the lift.
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Expert's answer
2012-09-06T09:59:45-0400
When the lift is at rest, a body produces an elongation of dl1 = 2 cm in the spring. When the lift moves downward with an acceleration, the elongation is changed to dl2 = 1.95 cm. dl = F * k dl1 = m * g * k dl2 = m * (g + a) * k dl1 / dl2 = g / (g + a) a = g * dl2 / dl1 - g a = 9.81 * 1.95 / 2.00 - 9.81 = -0.24525 So, the acceleration of the lift is a = -0.24525 m/s^2.
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