A 0.20 kg object, attached to a spring with spring constant k = 10 N/m, is moving on a horizontal frictionless surface in simple harmonic motion of amplitude of 0.080 m. What is its speed at the instant when its displacement is 0.040 m?

The speed is 49 cm/s

Explanation:

Given;

mass of the object, m = 0.20 kg

spring constant, k =  10 N/m

amplitude, A = 0.080 m

Total energy of the spring = ¹/₂kA² = ¹/₂ x 10 x (0.08)²

Total energy of the spring = 0.032 J

When its displacement is 0.040 m = half of Amplitude

Applying conservation of energy

Total energy of the spring = PE + KE

0.032 = ¹/₂kA² + ¹/₂mv²

0.032 =  ¹/₂ x 10 x (0.04)² +  ¹/₂ x 0.2 x v²

0.032 = 0.008 + 0.1v²

0.1v² = 0.032 - 0.008

0.1v² = 0.024

v² = 0.24

v = √0.24

v = 0.49 m/s = 49 cm/s

Therefore, the speed is 49 cm/s

the answer would be a

c) up is the direction the magnetic force is acting on the charge.