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Q. A 2 kg block is dropped from a height of 0.4m on a spring of force constant k=1960nm^-1.the maximum compression of the spring is?
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Q. A 2 kg block is dropped from a height of 0.4m on a spring of force ...
Mass of the body is ( m ) = 2 Kg
spring's constant ( K ) = 1960 N/m 
height of the body = 0.4 m
Let the maximum distance spring will compressed = x m
see the attachment , it is clear that 
initial potential energy of body = mg(h + x) 
final potential energy = spring potential energy = 1/2 Kx^2 
according to law of conservation of energy,
initial potential energy = final potential energy
mg(h + x) = 1/2 Kx^2
   2mgh + 2mgx = Kx^2
  Kx^2  - 2mgx - 2mgh = 0
 ⇒ 1960x^2 - 2*2*9.8*x - 2*2*9.8*0.4 = 0 
⇒ 1960x^2 - 39.2x - 15.68   = 0
 after solving this quadratic equations , we get x = 0.1m
     hence, answer is x = 0.1m 

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Q. A 2 kg block is dropped from a height of 0.4m on a spring of force ...
Problem: A 2 kg block is dropped from a height of 0.4m on a spring of force constant k=1960 N/m. The maximum compression of the spring is?

Solution:

Given data:
- Mass of the block (m) = 2 kg
- Height (h) = 0.4 m
- Force constant (k) = 1960 N/m

Step 1: Calculate the potential energy of the block

The potential energy (PE) of the block is given by the formula:

PE = mgh

where m is the mass, g is the acceleration due to gravity, and h is the height.

Given: m = 2 kg, g = 9.8 m/s², h = 0.4 m

PE = (2 kg) x (9.8 m/s²) x (0.4 m)
= 7.84 J

Step 2: Calculate the maximum potential energy converted to spring potential energy

The maximum potential energy converted to spring potential energy is given by the formula:

PE_spring = (1/2) k x (x²)

where k is the force constant of the spring and x is the maximum compression of the spring.

Given: k = 1960 N/m

We need to find x.

Step 3: Equating the potential energy of the block to the spring potential energy

Since the potential energy of the block is converted to spring potential energy, we can equate the two equations:

PE = PE_spring

7.84 J = (1/2) (1960 N/m) (x²)

Simplifying:

7.84 J = 980 N/m (x²)

Dividing both sides by 980 N/m:

0.008 = x²

Taking the square root of both sides:

x = √(0.008) m
= 0.089 m

Therefore, the maximum compression of the spring is 0.089 m.
Community Answer
Q. A 2 kg block is dropped from a height of 0.4m on a spring of force ...
9.1 cm
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