A 2.0- kg block is dropped from a certain height above a spring of spring consta

Question

A 2.0- kg block is dropped from a certain height above a spring of spring constant k- 1900 N/m that is in its relaxed position. Just before the block strikes the spring, it has speed v. See figure 1 The spring is then compressed a maximum distance d 0.10 m before the block comes momentarily to rest. (a) ( 15 points) What is the speed v of the block just before the block strikes the spring? See figure1 (b) (5 points) From what height h above the spring was the block dropped? See figure 2 at rest tigure 1 h-?

Explanation / Answer

total mechanical energy just before hitting the spring

Ei = m*g*d + (1/2)*m*v^2

total mechanical energy s=after the block momentarily comes to rest

Ef = (1/2)*k*d^2

from energy conservation

total energy is constant at any point

Ef = Ei

(1/2)*k*d^2 = m*g*d + (1/2)*m*v^2

(1/2)*1900*0.1^2 = (2*9.8*0.1) + (1/2)*2*v^2

speed v = 2.745 m/s

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along vertical

initial velocity voy = 0

final velocity vy = v

acceleration ay = -g = -9.8 m/s^2

displacement y = -h

from equation of motion

vy^2 - voy^2 = 2*ay*y

2.745^2 - 0 = 2*9.8*h

height h = 0.384 m

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