Signboards on Indian highways warn that speed thrills but kills. Driving instructors add a rule that surprises most learners. A car at \(72\,\text{km/h}\) does not need twice the braking distance of a car at \(36\,\text{km/h}\). It needs four times as much. Traffic police use the same rule in reverse. From the length of the skid marks, they calculate how fast a car was going.
The energy of motion
A moving body can do work. A moving hammer drives a nail into wood, and a moving cricket ball pushes a fielder's hands back. So a moving body carries energy, called kinetic energy. How much? Let a constant net force \(F\) act on a body of mass \(m\), from rest, along a straight line. Find its work by the time the speed is \(v\). The acceleration is \(a = F/m\), and over a distance \(s\), \(v^2 = 2as\). So the work is \(Fs = mas = \tfrac12 mv^2\).
What kind of quantity it is
Kinetic energy depends on the speed only, not on the direction of motion. It is a scalar, and since \(v^2\) is never negative, it is never negative. Two cars that approach each other head-on both carry positive kinetic energy; the two amounts do not cancel. Kinetic energy is proportional to the mass and to the square of the speed. A cricket ball of \(0.16\,\text{kg}\) bowled at \(40\,\text{m/s}\) carries \(\tfrac12 \times 0.16 \times 1600 = 128\,\text{J}\). A car of \(1000\,\text{kg}\) at \(20\,\text{m/s}\), which is \(72\,\text{km/h}\), carries \(200\,000\,\text{J}\).
Kinetic energy can also be written with the momentum \(p = mv\): \(K = p^2/2m\). For the same momentum, the lighter body has the larger kinetic energy. That is the skaters of Lesson 4 again, and the rifle and its bullet.
The speed of a car doubles. By what factor does its kinetic energy change?
Rank these bodies by their kinetic energy, largest first.
From forces to speeds in one step
Let a constant net force act on a body along a straight line. Its speed changes from \(u\) to \(v\) over a displacement \(s\). Newton's second law and the third equation of motion link the work of the net force to the two speeds.
\[ F_{\text{net}} = ma \]
Why. Newton's second law along the line of motion.
\[ W_{\text{net}} = F_{\text{net}}\,s = mas \]
Why. the net force is constant and points along the displacement.
\[ as = \tfrac12\left(v^2 - u^2\right) \]
Why. from the third equation of motion, \(v^2 = u^2 + 2as\).
\[ W_{\text{net}} = \tfrac12 mv^2 - \tfrac12 mu^2 \]
Read it. substitute step 3 into step 2. The acceleration and the time have gone.
The net work done on a body by all the forces acting on it equals the change in its kinetic energy: \(W_{\text{net}} = K_f - K_i\). The theorem holds for constant and for variable forces, along straight and curved paths.
The sign of the net work tells the story. Positive net work raises the kinetic energy, and the body speeds up. Negative net work lowers it, and the body slows down. Zero net work leaves the speed unchanged, even when large forces act, as for a box dragged across a floor at a steady speed. The theorem never asks for the acceleration or the time. It links forces and distances directly to speeds.
Four numbers from earlier lessons
The theorem finishes four stories left open. The student in the lift of Lesson 3 received a net work of \(400\,\text{J}\), so \(\tfrac12 \times 50 \times v^2 = 400\) and \(v = 4\,\text{m/s}\). The Atwood pair of \(5\,\text{kg}\) received \(10\,\text{J}\), so \(v = 2\,\text{m/s}\). The \(20\,\text{kg}\) crate on the speeding bed received \(250\,\text{J}\) from friction, so \(v = 5\,\text{m/s}\), the bed's own speed. In Lesson 4 the block had \(\tfrac12 \times 2 \times 6^2 = 36\,\text{J}\), and the block and plank together end with \(\tfrac12 \times 6 \times 2^2 = 12\,\text{J}\). The missing \(24\,\text{J}\) is the net work of the friction pair.
The derivation above used a constant force. The theorem is more general, and calculus proves it for a force that changes with position. Fill in the missing line.
\[ F = m\dfrac{dv}{dt} \]
Why. Newton's second law for a force that may change from place to place.
?
Your turn. the chain rule, \(\dfrac{dv}{dt} = \dfrac{dv}{dx}\dfrac{dx}{dt}\), with \(\dfrac{dx}{dt} = v\). Which line is it?
\[ \int_{x_1}^{x_2} F\,dx = \int_u^v mv\,dv \]
Why. multiply both sides by \(dx\) and integrate. As \(x\) goes from \(x_1\) to \(x_2\), the speed goes from \(u\) to \(v\).
\[ \int_{x_1}^{x_2} F\,dx = \tfrac12 mv^2 - \tfrac12 mu^2 \]
Read it. the integral of \(v\,dv\) is \(\tfrac12 v^2\). The left side is the work of the variable force.
A \(2\,\text{kg}\) trolley moves at \(3\,\text{m/s}\) on a smooth level floor. A steady push of \(10\,\text{N}\) acts on it along its motion for \(4\,\text{m}\). Find its final speed, in m/s.
Add every work
The theorem needs the net work: the sum of the works of all the forces on the body. The commonest mistake is to use only the work of the force that the question names. So list every force, write the work of each, and add them with their signs. Write the zero works down too, so that you know you have checked them.
In Fig 5.1 the pull's bar grows three times as fast as the friction bar falls. The kinetic energy bar always reaches the dashed line, the pull's work minus the friction's. The weight and the normal force act on the block too. They are at right angles to the motion and do no work, so they have no bar. After \(5\,\text{m}\) the block has \(100\,\text{J}\), and \(\tfrac12 \times 2 \times v^2 = 100\) gives \(v = 10\,\text{m/s}\). The acceleration was never needed.
A \(2\,\text{kg}\) block starts from rest and slides \(12.5\,\text{m}\) down a rough slope inclined at \(37°\). The coefficient of kinetic friction is \(0.25\). Take \(g = 10\,\text{m/s}^2\), \(\sin 37° = 0.6\) and \(\cos 37° = 0.8\). Find its speed at the bottom.
Three forces act: gravity, the normal force and kinetic friction. The block drops \(12.5 \times 0.6 = 7.5\,\text{m}\). The normal force is \(N = mg\cos 37° = 16\,\text{N}\), so friction is \(0.25 \times 16 = 4\,\text{N}\), up the slope.
\(W_g + W_N + W_f = \tfrac12 mv^2 - 0\).
\(W_g = mgh = 20 \times 7.5 = 150\,\text{J}\). \(W_N = 0\). \(W_f = -4 \times 12.5 = -50\,\text{J}\). So \(\tfrac12 \times 2 \times v^2 = 100\,\text{J}\), and \(v^2 = 100\).
The speed at the bottom is \(10\,\text{m/s}\). On a smooth slope it would be \(\sqrt{150} \approx 12.2\,\text{m/s}\): friction took one third of gravity's work.
In an accelerating frame
The theorem holds in an inertial frame, where Newton's laws hold. In an accelerating frame, such as the inside of a truck that speeds up, Newton's second law works only with a pseudo force \(-m\vec a_0\) added. The work-energy theorem works there too, provided the pseudo force's work is added to the net work. Speeds, displacements and works all change from frame to frame. The theorem holds in each frame, with that frame's numbers.
A \(2\,\text{kg}\) block rests on the smooth floor of a truck. The truck starts from rest with an acceleration of \(3\,\text{m/s}^2\). Using the truck's frame, find the speed of the block relative to the truck after it has slid \(6\,\text{m}\) towards the back.
In the truck's frame a pseudo force \(ma_0 = 2 \times 3 = 6\,\text{N}\) acts on the block, towards the back. The block slides \(6\,\text{m}\) that way, so the pseudo force does \(6 \times 6 = 36\,\text{J}\).
The weight and the normal force are vertical and do no work, and the floor is smooth. So \(\tfrac12 \times 2 \times v^2 = 36\), and \(v = 6\,\text{m/s}\) towards the back of the truck.
Check from the ground: the block stays at rest, and in \(2\,\text{s}\) the truck covers \(6\,\text{m}\) and reaches \(6\,\text{m/s}\). So, seen from the truck, the block moves back at \(6\,\text{m/s}\).
A block is dragged across a rough floor at a perfectly steady speed by a horizontal rope. What is the net work done on the block?
Stopping distance
A car of mass \(m\) moving at speed \(v\) brakes with a steady force \(F\). The braking force points against the motion, so its work is \(-Fs\) over a stopping distance \(s\). The car ends at rest. The theorem gives \(-Fs = 0 - \tfrac12 mv^2\), so \(s = mv^2/2F\). For braking by road friction, \(F = \mu mg\), and the mass cancels: \(s = v^2/2\mu g\). The stopping distance grows as the square of the speed.
Two cars brake equally hard on the same dry road. Car A was moving at \(36\,\text{km/h}\) and stops in \(10\,\text{m}\). Car B was moving at \(72\,\text{km/h}\). How far does car B travel after its brakes are applied?
In the figure the road gives a deceleration of \(5\,\text{m/s}^2\), which means \(\mu = 0.5\). Car A at \(10\,\text{m/s}\) stops in \(2\,\text{s}\) and \(10\,\text{m}\). Car B at \(20\,\text{m/s}\) takes twice the time, \(4\,\text{s}\), at twice the average speed, so it covers four times the distance, \(40\,\text{m}\). A driver also needs about a second to react before braking, which adds a distance proportional to the speed.
Now return to the highway signs. At \(72\,\text{km/h}\) a car has four times its kinetic energy at \(36\,\text{km/h}\). Its brakes need four times the road to remove it. The traffic police read the rule backwards. A skid of \(40\,\text{m}\) on a road with \(\mu = 0.5\) gives \(v^2 = 2\mu g s = 2 \times 0.5 \times 10 \times 40 = 400\). So the car was moving at \(20\,\text{m/s}\), that is \(72\,\text{km/h}\).
A car moving at \(54\,\text{km/h}\) brakes on a road where \(\mu = 0.75\), and its wheels lock. Take \(g = 10\,\text{m/s}^2\). Find the stopping distance, in metres.
In 1722 the Dutch professor Willem 's Gravesande dropped brass balls into soft clay from different heights. A ball that hit the clay at twice the speed sank four times as deep, not twice. Émilie du Châtelet, working in France, read of these experiments. In her book Institutions de physique of 1740 she argued that the "force" of a moving body goes as \(mv^2\), not as \(mv\). The dent in the clay is a stopping distance, and it grows as the square of the speed.
A \(60\,\text{kg}\) passenger in a car at \(36\,\text{km/h}\) carries \(\tfrac12 \times 60 \times 10^2 = 3000\,\text{J}\) of kinetic energy. In a crash, a force must remove all of it. If the seat belt and the crumpling front of the car let the passenger stop over \(0.6\,\text{m}\), the average force is \(3000/0.6 = 5000\,\text{N}\). Stopped by a hard dashboard in \(5\,\text{cm}\), the force would be \(60\,000\,\text{N}\). The work is fixed, so a longer stopping distance means a smaller force.
- Using the work of one force in the work-energy theorem.The theorem needs the net work, the sum of the works of every force on the body.
- Treating kinetic energy as a vector, or letting it cancel between bodies moving opposite ways.Kinetic energy is a scalar and is never negative. The kinetic energies of different bodies add.
- Expecting twice the speed to need twice the stopping distance.Kinetic energy goes as \(v^2\), so against the same force the stopping distance also goes as \(v^2\).
- Forgetting the pseudo force when working in an accelerating frame.In an accelerating frame, add the work of the pseudo force \(-m\vec a_0\) to the net work.