Question

A 1.40 kg block is attached to a spring with spring constant 16.5 N/m. While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 42.0 cm/s.(a) What is the amplitude of the subsequent oscillations?(b) What is the block's speed at the point where x = 0.300 A?

Answer 1

Answer:

0.12234 m

0.40065 m/s

Explanation:

m = Mass of block = 1.4 kg

k = Spring constant = 16.5 N/m

x = Compresion of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

$\frac{1}{2}mv^2=\frac{1}{2}kA^2\\\Rightarrow A=\sqrt{\frac{mv^2}{k}}\\\Rightarrow A=\sqrt{\frac{1.4\times 0.42^2}{16.5}}\\\Rightarrow A=0.12234\ m$

Amplitude of the oscillations is 0.12234 m

Again, as the energy of the system is conserved we have

$\frac{1}{2}kA^2=\frac{1}{2}mv^2+\frac{1}{2}kx^2\\\Rightarrow v=\sqrt{\frac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\frac{16.5(0.12234^2-(0.3\times 0.12234)^2)}{1.4}}\\\Rightarrow v=0.40065\ m/s$

The block's speed at the point is 0.40065 m/s