Question

A 880·kg car strikes a huge spring at a speed of 16·m/s, compressing it 11.0·m. (a) What is the spring constant of the spring? N/m (b) How long does the spring take to stop the car? s

Answer 1

Part A.

Using energy conservation:

KEi + PEi = KEf + PEf

KEi = (1/2)*m*Vi^2

PEi = 0, since initially no compression

KEf = 0, since final speed of car is zero

PEf = (1/2)*k*x^2

So,

(1/2)*m*Vi^2 + 0 = 0 + (1/2)*k*x^2

k = m*(Vi/x)^2

k = 880*(16/11.0)^2

k = spring constant = 1861.8 N/m

Part B.

Time taken by spring to stop the car will be equal to one-fourth of the time period of spring, since car moves from equilibrium position to max compression point, So it takes T/4 time, So

T = 2*pi*sqrt (m/k)

t = time taken by car to stop = T/4 = (pi/2)*sqrt (m/k)

t = (pi/2)*sqrt (880/1861.8)

t = 1.08 sec

Please Upvote.