Question

A 950-kg car strikes a huge spring at a speed of 22m/s (fig. 11-54), compressing the spring 5.0m. (a) what is the spring stiffness constant of the spring? (b) how long is the car in contact with the spring before it bounces off in the opposite direction?

Answer 1

(a) The spring stiffness constant of the spring is 18,392 N/m.

(b) The time the car was in contact with the spring before it bounces off in the opposite direction is 0.23 s.

Kinetic energy of the car

The kinetic energy of the car is calculated as follows;

K.E = ¹/₂mv²

K.E = ¹/₂ x 950 x 22²

K.E = 229,900 J

Stiffness constant of the spring

The stiffness constant of the spring is calculated as follows;

K.E =  U = ¹/₂kx²

k = 2U/x²

k = (2 x 229,900)/(5)²

k = 18,392 N/m

Force exerted on the spring

F = kx

F = 18,392 x 5

F = 91,960 N

Time of impact

F = mv/t

t = mv/F

t = (950 x 22)/(91960)

t = 0.23 s

Learn more about spring constant here: brainly.com/question/1968517

#SPJ4