Question

A 0.16-kg block on a horizontal frictionless surface is attached to an ideal spring whose force constant (spring constant) is 360 N/m. The block is pulled from its equilibrium position at x = 0.000 m to a position x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the position is x = - 0.040 m. what is the speed of the block?

Answer 1

The speed of the block when the position is x = - 0.040 m is 1.9 m/s.

How to calculate the speed of the block?

The speed of the block is calculated by applying the principle of conservation of energy as follows.

K.E = U

¹/₂mv² = ¹/₂kx²

mv² = kx²

v² = kx² / m

v = √ (kx² / m)

where;

• k is the spring constant
• x is the extension of the spring
• m is the mass of the block

The given parameters include;

spring constant, k = 360 N/m

mass of the spring, m = 0.16 kg

extension of the spring, x = -0.04 m

v = √ (360 x (-0.04)² / 0.16)

v = 1.9 m/s