A bus climbs from the plains to a hill station by a long ghat road with many hairpin bends. A ropeway climbs the same height straight up the hillside, over a far shorter distance. The road is many times longer. For the same passenger, does gravity do more work on her on the road or on the ropeway?
One question for every force
Every force on a body can be asked the same question: what is its component along the displacement? A component along the motion means positive work. A component against the motion means negative work. No component along the motion means no work. This lesson asks that question of gravity, the normal force, friction and tension in turn.
Gravity: only the height matters
Near the ground, gravity is a constant force \(mg\), straight down. Let a body move in a straight line from a height \(h_1\) to a height \(h_2\). The component of its displacement along the force is the drop, \(h_1 - h_2\). So the work done by gravity is \(W_g = mg(h_1 - h_2) = -mg\,\Delta h\), where \(\Delta h = h_2 - h_1\) is the rise. A body that goes up has \(\Delta h > 0\), and gravity does negative work on it. A body that comes down gets positive work from gravity. A body that moves on the level gets none.
\[ W_g = mg\,s\cos(90^\circ - \theta) \]
Why. a block slides a length \(s\) down a smooth slope of angle \(\theta\). Its weight points straight down, at \(90^\circ - \theta\) to the direction of the slope.
\[ W_g = mg\,s\sin\theta \]
Why. \(\cos(90^\circ - \theta) = \sin\theta\). This is the component \(mg\sin\theta\) along the slope times the length \(s\).
\[ W_g = mgh \]
Read it. the height of the slope is \(h = s\sin\theta\). The angle and the length have disappeared.
Any path at all
A curved path can be cut into many short straight steps. On each step, gravity does \(mg\) times the drop of that step. Adding the steps adds the drops, and the drops add up to the total drop. So the work of gravity between two points depends only on their heights. It does not depend on the length of the path or on its shape.
Now return to the two ways up the hill. Suppose both climb \(1500\,\text{m}\). For a passenger of mass \(60\,\text{kg}\), gravity does \(-60 \times 10 \times 1500 = -900\,000\,\text{J}\), that is \(-900\,\text{kJ}\), on either route. The long road spreads the climb over a longer distance, so the bus needs a smaller force along the road. The work of gravity is the same on both.
Three children of equal mass slide from the same platform to the same ground. One uses a straight slide, one a curved slide and one a spiral slide. On which child does gravity do the most work?
A \(2\,\text{kg}\) book is lifted from the floor to a shelf \(1.5\,\text{m}\) higher. Take \(g = 10\,\text{m/s}^2\). Find the work done by gravity on the book, in joules, with its sign.
The normal force: usually zero
The normal force is perpendicular to the surface of contact. A body sliding over a fixed surface, flat or sloping, moves along the surface. So the normal force is at right angles to the displacement at every instant, and its work is zero. This is why the normal force plays no part in most energy calculations.
When the surface moves
The rule "the normal force does no work" is a habit, not a law. It holds only while the surface does not move along its own normal. Stand on the floor of a lift that goes up. The floor pushes you up and you move up with it, so the normal force does positive work on you. In a lift going down, the floor still pushes up while you move down, so it does negative work.
A \(50\,\text{kg}\) student stands in a lift that accelerates upward at \(2\,\text{m/s}^2\) from rest, and rises \(4\,\text{m}\). Take \(g = 10\,\text{m/s}^2\). Find (a) the normal force of the floor, (b) its work on the student, (c) the work of gravity, (d) the total work.
(a) Newton's second law upward: \(N - mg = ma\), so \(N = m(g + a) = 50 \times 12 = 600\,\text{N}\).
(b) The floor pushes up and the student moves up \(4\,\text{m}\): \(W_N = 600 \times 4 = +2400\,\text{J}\).
(c) \(W_g = -mg\,\Delta h = -500 \times 4 = -2000\,\text{J}\).
(d) The total is \(2400 - 2000 = +400\,\text{J}\). It is positive, and the student speeds up. Lesson 5 turns this number into a speed.
You stand on the floor of a lift that is moving up. What work does the normal force of the floor do on you?
Kinetic friction on a fixed floor: always negative
Kinetic friction on a body sliding over a fixed surface points against the body's velocity. So the angle between friction and each small step is \(180°\), and every step gives negative work. For a friction of constant size \(f_k = \mu_k N\), the work is \(-f_k\) times the length of the path actually slid. A box slid \(3\,\text{m}\) out and \(3\,\text{m}\) back loses energy on both legs. Friction does \(-f_k \times 6\,\text{m}\) on it, not zero.
Static friction can do positive work
Friction does not always take energy away. Put a crate on the flat bed of a truck and let the truck speed up. The crate speeds up with the truck without slipping. The only horizontal force on the crate is the static friction from the bed, so that friction must point forward. The crate also moves forward. Force and displacement agree, so static friction does positive work on the crate. It is the force that carries energy from the engine into the crate.
Play both chips in Fig 3.2. The crate's velocity points forward in both, and so does its displacement. Only the friction arrow turns round. When the bed speeds up, friction is the force that speeds up the crate, and its work is positive. When the bed brakes, friction slows the crate, and its work is negative. The sign of the work follows the direction of the force, not the word "friction".
"Friction does negative work" is true for kinetic friction on a body sliding over a fixed surface. It is not a rule for every friction. Friction opposes the sliding of one surface over the other, not the motion of the body over the ground. Find the direction of friction first, then compare it with the body's displacement.
A \(5\,\text{kg}\) block is pulled \(2\,\text{m}\) up a rough ramp inclined at \(37°\) by a force of \(50\,\text{N}\) along the ramp. 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 the work done by each force, and the total.
The pull points along the displacement: \(W_F = 50 \times 2 = +100\,\text{J}\).
The block rises \(2 \times 0.6 = 1.2\,\text{m}\), so \(W_g = -mg\,\Delta h = -50 \times 1.2 = -60\,\text{J}\).
The normal force is at right angles to the ramp: \(W_N = 0\). It is \(N = mg\cos 37° = 40\,\text{N}\).
Friction is \(0.25 \times 40 = 10\,\text{N}\), down the ramp: \(W_f = -10 \times 2 = -20\,\text{J}\).
The total is \(100 - 60 + 0 - 20 = +20\,\text{J}\).
A crate sits on the flat bed of a truck without slipping while the truck speeds up. What work does friction do on the crate?
A \(10\,\text{kg}\) box is slid \(4\,\text{m}\) across a floor and then \(4\,\text{m}\) back to where it started. The coefficient of kinetic friction is \(0.3\). Take \(g = 10\,\text{m/s}^2\). Find the total work done by friction, in joules, with its sign.
Tension over a fixed pulley
A light, inextensible string passes over a fixed, light pulley and joins two blocks, as in the Atwood machine. The tension \(T\) is the same at both ends. When one block rises by \(s\), the other falls by \(s\), because the string does not stretch. On the rising block, tension pulls up and the block moves up, so tension does \(+Ts\). On the falling block, tension pulls up and the block moves down, so tension does \(-Ts\). The two works cancel exactly. The string only passes energy from one block to the other.
Blocks of \(3\,\text{kg}\) and \(2\,\text{kg}\) hang from a light string over a fixed, light, smooth pulley. They are released from rest. Take \(g = 10\,\text{m/s}^2\). Find the work of every force on each block while the blocks move \(1\,\text{m}\), and the total.
The acceleration is \(a = (m_1 - m_2)g/(m_1 + m_2) = 10/5 = 2\,\text{m/s}^2\). The tension is \(T = m_2(g + a) = 2 \times 12 = 24\,\text{N}\).
Each work is force times displacement times \(\cos\theta\). The \(3\,\text{kg}\) block moves down \(1\,\text{m}\); the \(2\,\text{kg}\) block moves up \(1\,\text{m}\).
Tension: \(+24\,\text{J}\) on the \(2\,\text{kg}\) block and \(-24\,\text{J}\) on the \(3\,\text{kg}\) block. Gravity: \(+30\,\text{J}\) on the \(3\,\text{kg}\) block and \(-20\,\text{J}\) on the \(2\,\text{kg}\) block.
The two tension works add to zero. The total work on the pair is \(30 - 20 = +10\,\text{J}\), all of it from gravity. The blocks speed up, and in Lesson 5 this \(10\,\text{J}\) becomes their speed.
Two blocks hang from a light, inextensible string over a fixed, light pulley. What is the total work done by the tension on the two blocks, as they move?
A sloping belt at an airport, or a conveyor that lifts coal at a power station, raises its load by static friction. The belt's friction points up the slope, and the load moves up the slope, so friction does positive work on the load. At a steady speed gravity does an equal negative work, and the load's speed does not change. The motor that drives the belt supplies the energy, and friction carries it into the load.
- Using the path length in the work of gravity.Gravity's work is \(-mg\,\Delta h\). Only the change in height counts, never the length of the path.
- Assuming that the normal force never does work.It does no work only while the surface does not move along its normal. The floor of a moving lift does work.
- Using the displacement, not the distance, for kinetic friction on a round trip.Kinetic friction opposes the sliding on every leg. Its work is \(-f_k\) times the total path length.
- Assuming that friction always does negative work.Static friction on a crate carried by an accelerating truck does positive work. Compare the direction of friction with the displacement.