A500-gram mass is attached to a spring and executes simple harmonic motion with a period of 0.25 second. if the total energy of the system is 4j, find the force constant of the spring?

315.5 N/m

Explanation:

m = 500 g = 0.5 kg

T = 0.25 second

Total energy, E = 4 J

Let K be the spring constant.

The formula for the time period is given by

K = 315.5 N/m

work: w = fd (given)

work done by chet   =50*1 = 50 j

work done by mika =40*2 = 80 j

work done by sara =30*1.5 = 45 j

work done by bill =60*0.5 = 30 j

students in order from the greatest amount of work done to the least is mika> chet> sara> bill

hence option (b) is correct.