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Spring Force :Class 11 CBSE Physics Chapter 5

class 11 physics spring force

Spring Force:Class 11 CBSE Physics Chapter 5

The spring force is not so different from the force of string(Tension),here both of the mass of string and spring are supposed as massless,the spring force(F) is directly proportional to the displacement(Δx) in the direction of force.The spring force is the same if it is stretched or compressed by the same magnitude but its direction is always opposite to the direction displacement.

class 11 physics spring force

 

Let a spring is stretched by a force FSP  to a displacement of Δx

The spring force is directly proportional to the displacement Δx

FSP  ∝ Δx

The minus sign shows that the direction of spring force is opposite of the displacement or elongation

∴FSP  = -kΔx

Where k is the constant force of the spring and depends on the type of spring,the spring force is a variable force,it increases with the increase of elongation Δx.

spring force(Hook's Law)

If the spring is compressed then the spring force will be in opposite direction.The spring force, FSP is same at every point of the spring in case of the same spring.

If a spring is pulled by two persons,the displacement of the spring in both sides are x1 and x2

Then net elongation is

Δx = x1 + x2

The spring force will be the same at every point of the spring and its value is

FSP = -k(x1 + x2)

In case of spring and string the spring force and the force of tension have the same value at every corresponding points of them when both are considered mass less.

Example: Two blocks A and B are attached to the same spring ,k=50N/m. Then find

(a)The spring force when blocks A and B are displaced by 1 m on both sides.

(b)The spring force when blocks A and B are displaced by 1 m in the same direction.

Solution. For both cases (a) and (b) the images are shown below.

(a)The displacement,of the block is 1m

Spring force is

FSP = -kΔx

Neglecting the minus sign since we are needed to evaluate only the magnitude of spring force

FSP = kΔx

Displacement ,Δx = 1 +1 = 2m

FSP = 50×2 = 100N

 

questions on The spring force

(b) The net displacement ,Δx = 1-1 =0

FSP = 50×0 = 0N

The calculation of the elongation when a weight is suspended on a spring :

Let a solid of the mass m is suspended on a spring and spring is displaced by Δx then the spring force is

FSP = kΔx

Since spring is supposed as it is in equilibrium condition

FSP = mg

mg = kΔx

Δx = mg/k

If mass m and value of k is given then we can calculate the displacement of the spring.

weight suspended on a spring

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