A 1200-kg car rolling on a horizontal surface has speed v = 65 km/h when it strikes a horiztonal coiled spring and is brought to rest in a distance of 2.2 m. What is the spring stiffness constant of the spring? How do you work out the formula?
2006-10-11 03:41:33 · 3 answers · asked by Pat M 1 in Science & Mathematics ➔ Physics
the formula for energy in a spring is
1/2kx^2
using conservation of energy, calculate the kinetic energy of the car
=1/2 m v^2
=1/2 *1200kg *
(65km/hr*60min/hr
*60sec/min*1000m/km)^2
the energy of the spring is equal to that
=1/2 *k *2.2^2
k=1.36x10^19
that's a strong spring
j
2006-10-11 03:50:14 · answer #1 · answered by odu83 7 · 0⤊ 4⤋
WEL STIFNESS OF ASTRING IS THE MESURE OF THE FORCE REQUIRED TO COMPRESS THE SPRING THU A CERTAIN LENGTH MORE IS THE STFNESS MORE IS THE FORCE REQ.
IN THE NUMERICAL WE CAN CNSERVE THE ENERGY BEFORE AND AFTER THETHE COLLISION
INI ENERGY IS 1/2MV^2
FINAL IS 1/2KX^2
HERE K IS THE STIFNESS OF THE THE STRING
K=MV^2/X^2
M=1200
V=65/3.6
X=2.2
2006-10-11 04:02:23 · answer #2 · answered by Anonymous · 0⤊ 3⤋
k= (mv^2)/x^2
2014-12-22 07:12:05 · answer #3 · answered by Daniela 2 · 0⤊ 1⤋