Energy in a spring:
E = 0.5 * k * x²
k spring constant = 800 n/m
x stretch of the spring = 5 cm = 0.05 m
E = 0.5 * 800 * 0.05² = 1
The mass m of the object = 5.25 kg
<h3>Further explanation</h3>
Given
k = spring constant = 3.5 N/cm
Δx= 30 cm - 15 cm = 15 cm
Required
the mass m
Solution
F=m.g
Hooke's Law
F = k.Δx

We can rearrange the mirror equation before plugging our values in.
1/p = 1/f - 1/q.
1/p = 1/10cm - 1/40cm
1/p = 4/40cm - 1/40cm = 3/40cm
40cm=3p <-- cross multiplication
13.33cm = p
Now that we have the value of p, we can plug it into the magnification equation.
M=-16/13.33=1.2
1.2=h'/8cm
9.6=h'
So the height of the image produced by the mirror is 9.6cm.
Given:
u(initial velocity)=0
a=5.54m/s^2
v(final velocity)=2 m/s
v=u +at
Where v is the final velocity.
u is the initial velocity
a is the acceleration.
t is the time
2=0+5.54t
t=2/5.54
t=0.36 sec