Question

A block of mass 2 kg initially at rest is dropped from a height of 1m into a vertical spring having force constant 490Nm-1. Calculate the maximum distance through which the spring will be compressed? (g=9.8ms-2)

Answer 1

Answer:

0.2 m

Explanation:

From conservation of energy

Potential energy of particle at height "h" is converted to potential energy stored in spring.

So,

mgh = 0.5kx²

From above

x = √[(mgh) / (0.5 k)]

= √[(1 kg × 9.8 m/s² × 1 m)/(0.5 × 490 N/m)]

= √(4/100) m

= 2/10 m

= 0.2 m

Answer 2

Given,  

Block mass = 2 Kg

Height of the body = 1 m

Spring constant = 490 N/m

Acceleration due to gravity = 9.8 m/s²

To find,  

Maximum distance through which the spring is going to be compressed

Solution,  

We can simply solve this numerical problem by using the subsequent mechanism

Block mass (m) = 2 kg

Height of the body (h)  = 1 m

Spring constant (K) = 490N/m

Acceleration due to gravity (g)  = 9.8m/s²

Now,

Let the maximum distance through which the springs is compressed = ‘X’

Initial potential energy of body = mg(h+X)

Final potential energy = Spring potential energy = 1/2Kx²

From  the law of conservation of energy,

Initial potential energy = Final potential energy

⇒ mg(h+X) = 1/2(Kx²)

⇒ 2mgh+2mgX = Kx²

⇒ Kx²−2mgX−2mgh = 0

⇒ (490X²)−(2×2×9.8×X)−(2×2×9.8×1) = 0

⇒ 490X²−39.2X²−39.2 = 0

After solving the above quadratic equations, we get x = 0.2m

Therefore, the maximum distance through which the springs is compressed is 0.2 m