Answer:
0.34 sec
Explanation:
Low point of spring ( length of stretched spring ) = 5.8 cm
midpoint of spring = 5.8 / 2 = 2.9 cm
Determine the oscillation period
at equilibrum condition
Kx = Mg
g= 9.8 m/s^2
x = 2.9 * 10^-2 m
k / m = 9.8 / ( 2.9 * 10^-2 ) = 337.93
note : w = =
= 18.38 rad/sec
Period of oscillation =
= 0.34 sec
Explanation:
Given that,
The initial velocity of a skater is, u = 5 m/s
She slows to a velocity of 2 m/s over a distance of 20 m.
We can find the acceleration of skater. It is equal to the rate of change of velocity. So, it can be calculated using third equation of motion as follows :
a = acceleration
So, her acceleration is and she is deaccelerating. Also, her initial velocity is given i.e. 5 m/s.