Question

Q11: a box hit a spring. the box is 5kg hitting the spring at 5m/s. The spring was 18cm(?) and 12cm(?) after the box hit and stopped. What is the spring constant? what if the box hit at 2.5m/s with same mass, how much would the spring move?

Answer 1

Mass of box ,m=5kg

Initial speed of box, v_i=5 m/s

Final speed of box, v_f=0 m/s

Compression in spring, x=18cm-12cm=6 cm=0.06 m

According to conservation of energy

Change in kinetic energy of box=Potential energy of spring

(1/2)mv_i² - (1/2)mv_f² =(1/2)Kx²

(1/2)mv_i²-0=(1/2)Kx²

Spring constant,K=mv_i²/x² ……………………….(1)

K=(55²)/(0.06²)=34722.22 N/m

Part b.

If v=2.5

Then from equation 1 is

x=(mv_i²/K)^1/2=((52.5²)/(34722.22))^1/2=0.03m

Spring is moved by=3 cm