Spring force is given by the expression :

F = - k x F = - k x

Figure 1: Spring force

Spring

where k is spring constant, measured experimentally for a particular spring. The value of k (=-F/x) measures the stiffness of the spring. A high value indicates that we would require to apply greater force to change length of the spring.

This relationship, defining spring force, is also known as “Hooke’s law”, which is valid under two specific conditions :

The attribute “x” is measured from the position of “neutral length” or “relaxed length” - the length of spring corresponding to situation when spring is neither stretched nor compressed. We shall term this position as origin.

The negative sign in the expression indicates an important aspect of the nature of spring force that it is always directed towards the position at neutral length. See the figure to visualize how spring force is always directed towards origin.

Figure 2: Spring force is always directed towards origin

Spring

Cotext of spring force

We must, here, understand that forces on the body (block) attached to spring may involve friction, if the surfaces between block and the surface is rough. For the sake of simplicity, we consider here that the surfaces are smooth. In this situation, we need not consider force due to gravity as it is normal to the motion and does not influence our consideration. As such, we will not consider gravity, while calculating work for the motion on smooth surface.

Just like the case of gravity, there are two situations with respect to application of force on a body by the spring.

1: In one case, the body is given a jerk, say towards right and the body is let go. In this case, the only force on the block is spring force. This situation is similar to vertical projection of an object with an initial speed. This case of spring force, therefore, is specified by an initial speed, preferably at the origin.

Figure 3: Only force on the block is spring force.

Spring

2: In other case, there is external force(s) in addition to spring force. The horizontal force is applied on the block as it moves along the surface. This situation is similar to the case of pulling a lift vertically by string and pulley arrangement. In this case, we would be required to analyze external force with the help of “free body diagram”.

Figure 4: There is other force (tension, T) besides spring force.

Spring

Now, we are set to obtain an expression for the work done by the spring on a block as shown in the figure. We give the block a jerk to the right as discussed earlier. Let the extension of the spring be “x”. Then,

F = - k x F = - k x

As the force is variable, we can not use the expression “Frcosθ” to determine work. It is valid for constant force only. We need to apply expression, involving integration to determine work :

W S = ∫ F ( x ) ⅆ x W S = ∫ F ( x ) ⅆ x

Let x i x i and x f x f be the initial and final positions of the block with respect to origin, then

W S = ∫ x i x f - k x ⅆ x ⇒ W S = - k [ x 2 2 ] ] x i x f ⇒ W S = 1 2 k ( x i 2 - x f 2 ) W S = ∫ x i x f - k x ⅆ x ⇒ W S = - k [ x 2 2 ] ] x i x f ⇒ W S = 1 2 k ( x i 2 - x f 2 )

If block is at origin, then x i = 0 x i = 0 and x f = x x f = x (say), then

⇒ W S = - 1 2 k x 2 ⇒ W S = - 1 2 k x 2

When block is subjected to the single force due to spring, the work is given by above two expressions. During motion towards right, the spring force on the block acts opposite to the direction of motion. It increases in magnitude with increasing displacement. Thus, spring force does negative work transferring energy "from" the block. As a consequence, kinetic energy of the block decreases.

This process continues till the velocity and kinetic energy of the block are zero. Spring force, then, pulls block towards the mean position i.e. in negative x - direction. The spring force, now, is in the direction of motion. It does the positive work on the block transferring energy to the block.

The process continues till the particle returns to the initial position, when its velocity is same as that in the beginning. As such kinetic energy of the block on return at mean position is equal to that in the beginning.

⇒ K f = K i ⇒ K f = K i

For the roundtrip, net work by spring force is zero. Net transfer of energy “to” or “from” the particle is zero. Initial kinetic energy of the particle is retained at the end of round trip. Thus, we can see that the motion under spring is lot similar as that of motion under gravity.

We must understand that work by spring force, for a given displacement, is independent of the presence of other forces. The work done by spring remains same.

For this situation, “work-kinetic energy” theorem has following form :

⇒ K f - K i = W S + W F ⇒ K f - K i = W S + W F

where W S W S and W F W F are the work done by the spring force and other applied force(s) respectively.

Here, work done by other external force(s) may be analyzed with respect to following different conditions :

In the first two cases, initial and final kinetic energy are same. Hence,

⇒ K f - K i = W S + W F = 0 ⇒ W F = - W S ⇒ K f - K i = W S + W F = 0 ⇒ W F = - W S

This is an important result. This means that we can simply compute the work done by spring force and assign the same preceded by a negative sign as the work done by other force(s). Such situation can arise when external force displaces the block and extends the spring to a certain extension such that end velocities are either zero or same.

In third case, kinetic energies at end points are not same. However, work done by spring force remains same as before. Thus, the difference in kinetic energy during a motion is attributed to the net work as done by spring and other forces.

Example 2

Problem : A block of 1 kg with a speed 1 m/s hits a spring placed horizontally as shown in the figure. If spring constant is 1000 N/m, find the compression in the spring.

Figure 5: Spring is compressed by the striking block.

Spring and block

Solution : When block hits the spring, it is compressed till the block stops. Here, we see that kinetic energies at the beginning and at the point when block stops are not same. Note that the only force acting on the block is due to spring. Hence,

Figure 6: At the end block comes to a stop momentarily.

Spring and block

K f - K i = W S K f - K i = W S

⇒ W S = K f - K i = 0 - 1 2 m v 2 ⇒ W S = - 0.5 x 1 x 1 2 = - 0.5 J ⇒ W S = K f - K i = 0 - 1 2 m v 2 ⇒ W S = - 0.5 x 1 x 1 2 = - 0.5 J

Now, work by spring is :

⇒ W S = - 1 2 k x 2 ⇒ W S = - 0.5 x 1000 x x 2 = - 500 x x 2 ⇒ W S = - 1 2 k x 2 ⇒ W S = - 0.5 x 1000 x x 2 = - 500 x x 2

Combining two values, we have :

⇒ - 250 x x 2 = - 0.5 ⇒ - 250 x x 2 = - 0.5

⇒ x 2 = 0.5 500 = 0.001 ⇒ x = 0.032 m ⇒ x 2 = 0.5 500 = 0.001 ⇒ x = 0.032 m