acceleration

2012-10-20 4:11 am

A 1.80kg box and a 3.30kg box on a perfectly smooth horizontal floor have a
spring of force constant 350 N/m compressed between them. The initial
compression of the spring is 5.00 cm.

Find the acceleration of 1.80 kg box the instant after they are released.

? m/s^2

更新1:

Find the acceleration of 3.30kg box the instant after they are released. m/s^2

回答 (1)

天同

2012-10-20 8:21 am

✔ 最佳答案

The total compression of 5 cm is shared between the two boxes inversely to their masses.
Hence, 1.8x = 3.3(5-x)
where x is the compression shared by the 1.8 kg box
solve for x gives x = 3.235 cm = 0.03235 m

Force acting on the 1.8kg box
= 350 x 0.03235 N
Acceleration = 350 x 0.03235 /1.8 m/s^2 = 6.29 m/s^2

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