A 1.2 kg block is dropped from a height of 0.5 m above an uncompressed spring. T

Question

A 1.2 kg block is dropped from a height of 0.5 m above an uncompressed spring. The spring has a spring constant k = 160 N/m and negligible mass. The block strikes the top end of the spring and sticks to it. a. Find the speed of the block when it strikes the top end of the spring. b. Find the period of oscillations of the block. c. Find the compression of the spring when the speed of the block reaches its maximum value. d. Find the maximum compression of the spring. e. Find the amplitude of oscillations of the block.

Explanation / Answer

(a)

The block falls with constant acceleration through a known distance v^2 =u^2 + 2ax

v^2=0^2 + 2(-9.8)(-0.5m) = 9.8

v= 3.13 m/s

(b)

This is a spring oscillating with a mass on it. It follows the motion equation which is the basis the mass-spring constant equation.

Tspring = 2 (m /k)^(1/2) = 0.544 sec

C)

The block will continue to accelerate (at a decreasing rate) because the spring initially applies a small upward force. As it is compressed, the spring applies more and more force. Eventually the force of gravity on it is exactly balanced by the force of the spring pushing up. After that position, the block will beging to slow down. The critical changing point happens when the net force is zero.

Fnet = 0 = –mg + kx= mg= kx

(1.2kg)(9.8) = (160)(x)

x = 0.0735 m

D).

The key here is that the initial gravitational potential energy is completely transformed momentarily into spring potential energy.

Ug = Uspring

mg(h+xmax) = ½ kx2

(2)(9.8)(0.5+x) =½ (160)*x^2

9.8+19.6x=80*x^2

80x^2 – 19.6x -9.8 = 0

x1=0.49 m

(and –0.25; the displacement updward from the equilibrium position)

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