A 1.37-kg block is held in place against the spring by a 74-N horizontal externa

ID: 3895421 • Letter: A

Question

A 1.37-kg block is held in place against

the spring by a 74-N horizontal external

force (see the figure). The external force is

removed, and the block is projected with a

velocity v1 upon separation from the

spring. The block then descends a ramp of

height h=2.7 cm and has a velocity v2 =

1.4 m/s at the bottom

Determine the initial compression distance of the spring and the spring constant of the spring.

x = 2.6 cm

k = 2790 N/m

Those are the answers, I would like an explanation to how they obtain those answers. Thank you.

veocity at the top of the ramp be V

using conservation of energy

.5mV^2 + mgh = .5m*v2^2

V = 1.2 m/s

energy stored in spring is converted to kinetic energy of the block

so .5kx^2 = .5mV^2 = .9864 J

kx = 74 N

.5kx^2 = .5x (kx) = .9864

.5x (74) = .9864

x = .0266 meter or = 2.6 cm aprox

k = 74/.0266 = 2790 N/m aprox

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