An engine is attached to a wagon through a shock absorber of length 1.5m. The system with a total mass of 50,000 kg is moving with a speed of 36 km h$^{-1}$ when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by 1.0 m. If 90% of energy of the wagon is lost due to friction, calculate the spring constant.

An engine is attached to a wagon through a shock absorber of length 1.5m. The system with a total mass of 50,0

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Concept understanding — Work Energy Theorem

The Work-Energy Theorem: From Intuition to Precision

Imagine pushing a heavy box across the floor. The harder you push and the farther it slides, the faster it moves when you let go. That connection — the push (force) over a distance (displacement) changing the box's speed — is exactly what the Work-Energy Theorem captures.

The Intuition First

Think of work as the "currency" that buys motion. When you do work on an object, you transfer energy to it. That energy shows up as kinetic energy — the energy of motion. The more work you do, the more the object's kinetic energy changes.

If you push a stationary ball, it starts moving. If you push a moving ball in the same direction, it speeds up. If you push against its motion, it slows down. In every case, the work done equals the change in the ball's kinetic energy.

Note

Work is done by a force on an object. The object's kinetic energy changes by exactly that amount (assuming no other forces do work).

The Precise Statement

W_{\text{net}} = \Delta K = K_f - K_i

Where:

$W_{\text{net}}$ is the net work done on the object (the total work from all forces combined)
$K_f$ is the final kinetic energy
$K_i$ is the initial kinetic energy

And kinetic energy is defined as:

K = \frac{1}{2}mv^2

So the theorem can also be written as:

W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

Why "Net" Work Matters

This is the most common point of confusion. The theorem uses net work — the work done by the net force (the vector sum of all forces). If you push a box and friction opposes it, the net work is the work you do minus the work friction does. Only that net amount changes the kinetic energy.

Watch out

If you push a box at constant speed, your work is positive, but friction does equal negative work. The net work is zero, so kinetic energy doesn't change — the box keeps moving at the same speed. Your work didn't "disappear"; it was dissipated as heat by friction.

A Simple Derivation (for constant force)

Consider a constant net force $F_{\text{net}}$ acting on an object of mass $m$ over a displacement $s$ . From Newton's second law:

F_{\text{net}} = ma

From kinematics (constant acceleration):

v_f^2 = v_i^2 + 2as

Multiply both sides by $\frac{1}{2}m$ :

\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + mas

But $mas = F_{\text{net}} s = W_{\text{net}}$ , so:

\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + W_{\text{net}}

Rearranging:

W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = \Delta K

Tip

The theorem holds even for variable forces and curved paths — the derivation uses calculus then, but the result is the same.

What It Tells You (and What It Doesn't) …

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The initial kinetic energy of the wagon is partially lost to friction and partially stored as potential energy in the shock absorber's spring. By calculating the initial kinetic energy, accounting for the 90% loss due to friction, and equating the remaining 10% to the spring's potential energy, we find the spring constant to be $\boxed{5 \times 10^5 \text{ N m}^{-1}}$ .

When a moving object is brought to rest, its kinetic energy must be transformed into other forms of energy. In this problem, the wagon's kinetic energy is converted into two main forms: heat due to friction and elastic potential energy stored in the shock absorber's spring.

The principle we use here is the Work-Energy Theorem, which states that the net work done on an object equals its change in kinetic energy. More broadly, we can think of it as the conservation of energy: the initial mechanical energy (kinetic energy in this case) is transformed into other forms.

KE_{initial} = E_{lost\_to\_friction} + PE_{spring}

Here, $KE_{initial}$ is the kinetic energy of the wagon before braking, $E_{lost\_to\_friction}$ is the energy dissipated as heat due to friction, and $PE_{spring}$ is the elastic potential energy stored in the compressed spring.

Let's break down the calculation step-by-step.

Convert the initial speed to standard units. The mass of the system is $m = 50,000 \text{ kg}$ . The initial speed is given as $v = 36 \text{ km h}^{-1}$ . To use this in energy calculations, we must convert it to metres per second ( $\text{m s}^{-1}$ ).

v = 36 \text{ km h}^{-1} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}}

v = 36 \times \frac{1000}{3600} \text{ m s}^{-1}

v = 10 \text{ m s}^{-1}

Calculate the initial kinetic energy of the system. The kinetic energy ( $KE$ ) of an object is given by the formula: $KE = \frac{1}{2}mv^2$

KE_{initial} = \frac{1}{2} \times 50,000 \text{ kg} \times (10 \text{ m s}^{-1})^2

KE_{initial} = \frac{1}{2} \times 50,000 \times 100 \text{ J}

KE_{initial} = 2,500,000 \text{ J} = 2.5 \times 10^6 \text{ J}

Determine the energy lost due to friction. The problem states that 90% of the energy of the wagon is lost due to friction. This means 90% of the initial kinetic energy is converted into heat.

E_{lost\_to\_friction} = 0.90 \times KE_{initial}

E_{lost\_to\_friction} = 0.90 \times 2.5 \times 10^6 \text{ J}

E_{lost\_to\_friction} = 2.25 \times 10^6 \text{ J}

Calculate the energy stored in the spring. Since 90% of the initial kinetic energy is lost to friction, the remaining energy must be stored in the spring as elastic potential energy. This remaining energy is $100\% - 90\% = 10\%$ of the initial kinetic energy. $PE_{spring} = 0.10 \times KE_{initial}$ …

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