A220 g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.150 s. if the total energy of the system is 1.50 j, find the following(a) the force constant of the spring(b) the amplitude of the motion.

a)K=385.86 N/m

b)A=0.088 m

Explanation:

Given that

m = 220 gm = 0.22 kg

t= 0.15 s

Total energy TE= 1.5 J

We know that time period in the SHM given as

ω=41.88 rad/s

We know that

ω² m = K

K=Spring constant

K=385.86 N/m

A=Amplitude

A=0.088 m

answer would be letter choice (a),   a ball resting on a shelf. would be another example of potential energy   >   stored energy > not moving.

a ball spinning on a finger   >   kinetic energy > moving energy

a ball rolling on the ground   >   kinetic energy   > moving energy

a   ball flying through the air   > kinetic energy   >   moving energy

hope that                                         : )

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s