Question

A small block of mass 100 g is pressed against a horizontal spring fixed at one end to compress the spring through 5.0 cm (Figure). The spring constant is 100 N/m. When released, the block moves horizontally till it leaves the spring. Where will it hit the ground 2 m below the spring?
Concept of Physics - 1 , HC VERMA , Chapter " Work and Energy"

Answer 1

Given in the question :- 
Mass , m = 100 g = 0.1 kg.
Spring compression , x = 5.0 cm = 0.05 m .
k = 100 N/m .

Now, stored potential energy in the spring = 
= 1/2 kx²
= (1/2) 100 × 0.05²
= 0.125 J.

At the time  body leaves the spring, let v be the velocity. Then the potential energy changes into Kinetic energy.
0.125= 1/2 mv².
0.125= (1/2) × 0.1 × v²
v² = 0.25/ 0.1
v² = 2.5
v = √2.5 = 1.58 m/s.

Let time take for the projectile to fall 2 m be t. 
In this direction u =0 and h = 2 
Now,  h = ut + 1/2gt²
 2 = 0 + (1/2) × 9.8 × t²
t² = 2/4.9
t² = √0.40
t = 0.63 sec
The block moves horizontally , so Now the horizontal distance :- 
 = v.t
=1.58 × 0.63
=1 m.



Hope it Helps.

Answer 2

hope it helps
ps. apologies fot my bad handwriting.