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Answer to Question #14137 in Mechanics | Relativity for jinan haneefc

Question #14137
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.
Expert's answer
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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