Question

A block of mass 4kg hangs from a spring of spring constant k = 400 N/m. The block is pulled down through 15 cm below and released. What is the kinetic energy when the block is 10 cm above the equilibrium position?

• A. 5J
• B. 2.5J
• C. 1J
• D. 1.9J

Answer 1

The correct option is B 2.5J

Amplitude = 0.15 m
$w=\sqrt{\frac{k}{m}}=\sqrt{\frac{400}{4}}=10rad/s\Rightarrow f=\frac{w}{2\pi }=\frac{5}{\pi }Hz$
Energy of the particle executing SHM = KE + PE
$\Rightarrow \frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2=\frac{1}{2}K{A}^2$
Therefore, $\frac{1}{2}m{v}^2=\frac{1}{2}k{A}^2-\frac{1}{2}k{x}^2=\frac{1}{2}(A^2-x^2)$
$=\frac{1}{2}\times 400({0.15}^2-{0.1}^2)=2.5J$