One end of a vertical spring of spring constant k = 1900 N/m is attached to the

Question

One end of a vertical spring of spring constant k = 1900 N/m is attached to the floor. You compress the spring so that it is 2.90 m shorter than its relaxed length, place a 1.00-kg ball on top of the free end, and then release the system at t = 0. (All values are measured in the Earth reference frame.)

1) By how much does the mass of the spring change during the time interval from t = 0 to the instant the ball leaves the spring?

mf-mi =

2) By how much does the mass of the Earth-ball-spring system change during the time interval from t = 0 to the instant the ball reaches its maximum height?

mf-mi =

Explanation / Answer

Here ,

spring constant , k = 1900 N/m

spring compressed , x = 2.90 m

mass of ball , m = 1 Kg

1)

as energy = 0.5 * k * x^2

(mf - mi) * c^2 = - 0.5 * 1900 * 2.90^2

(mf - mi) = - 8.88 *10^-14 Kg

te change in mass is -8.88 *10^-14 Kg

2)

Now , for the increase in mass of object

(mf - mi) * c^2 = same energy as the spring

(mf - mi) * c^2 = 0.5 * 1900 * 2.90^2

(mf - mi) = 8.88 *10^-14 Kg

the change in mass is 8.88 *10^-14 Kg

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.