Work
A cliff diver at La Quebrada climbs up the rocks and then leaps; a worker pushes a pallet in a warehouse; a cable car cabin goes up and down part of its route. In all three cases there are forces and there is displacement, but the energy that gets transferred depends on how the two combine. On this page you will learn to measure that transfer, called work, and to use it to predict speeds without needing the time.
Work as transferred energy
You push with all your might against a concrete wall for ten minutes and end up exhausted, but the wall doesn't move. On the other hand, a slight push on a skateboard sends it several meters. Why does physics say that in the first case you did no work and in the second you did?
Of the whole force, only the component in the direction of the displacement counts, $F\cos\alpha$. The perpendicular component, $F\sin\alpha$, does not push the object along its path and transfers no energy.
The sign indicates the direction of the transfer:
| Angle $\alpha$ | $\cos\alpha$ | Work | What happens |
|---|---|---|---|
| $0^\circ$ | $1$ | maximum positive | the force gives energy to the object |
| $90^\circ$ | $0$ | zero | no transfer (normal force, weight on a horizontal surface) |
| $180^\circ$ | $-1$ | negative | the force takes energy away (friction) |
The work done by a constant force on a moving body is $W = \vec{F} \cdot \vec{d} = F \cdot d \cdot \cos \alpha$, where $\alpha$ is the angle between the force and the…
The work done by a constant force on a moving body is $W = \vec{F} \cdot \vec{d} = F \cdot d \cdot \cos \alpha$, where $\alpha$ is the angle between the force and the displacement. Without performing calculations, state whether the work is positive, negative, or zero in each case, justifying your answer using the value of $\alpha$:
- a) The pulling force of a tractor towing a trailer along a horizontal path.
- b) The friction force opposing the motion of the trailer.
- c) The normal force exerted by the ground on the trailer.
- d) The force with which a worker holds a box at rest on their shoulder.
A warehouse worker pushes a pallet with a constant horizontal force of $250\ \text{N}$, moving it $12\ \text{m}$ along the floor in a straight line. a) What angle does…
A warehouse worker pushes a pallet with a constant horizontal force of $250\ \text{N}$, moving it $12\ \text{m}$ along the floor in a straight line.
- a) What angle does the force make with the displacement?
- b) Calculate the work done by the worker.
A person walks their dog, pulling on the lead with a constant force of $80\ \text{N}$ at an angle of $60^\circ$ to the horizontal over a straight section of…
A person walks their dog, pulling on the lead with a constant force of $80\ \text{N}$ at an angle of $60^\circ$ to the horizontal over a straight section of $25\ \text{m}$ ($\cos 60^\circ = 0.5$), as shown in the figure.
Calculate the work done.
Friction and the work done by each force
When several forces act on an object, each one has its own work. You calculate them one by one, using the angle of each force relative to the displacement, and then add them up: the result is the net work.
How much is the friction?
Often the problem does not give $f$, but the coefficient of kinetic friction $\mu$, a number without units that depends on the two surfaces in contact (rubber on dry asphalt: large $\mu$; puck on ice: small $\mu$).
A traveller pushes their luggage trolley through an airport terminal in a straight line on a horizontal floor. a) Reason, using the expression…
A traveller pushes their luggage trolley through an airport terminal in a straight line on a horizontal floor.
- a) Reason, using the expression $W = F \times d \times \cos \alpha$, why the work done by the normal force and the work done by the weight are zero during the journey.
- b) If the traveller exerts a horizontal force of $18\ \text{N}$ over $100\ \text{m}$, what work do they do?
When dragging a sofa $8\ \text{m}$ across a living room floor, the kinetic friction force is $35\ \text{N}$. a) What angle does the friction force make with the…
When dragging a sofa $8\ \text{m}$ across a living room floor, the kinetic friction force is $35\ \text{N}$.
- a) What angle does the friction force make with the displacement?
- b) Calculate the work done by friction and interpret its sign.
In a treasure hunt, a student drags a box of books $15\ \text{m}$ across the horizontal floor of the gymnasium by pushing with a horizontal force of $120\ \text{N}$,…
In a treasure hunt, a student drags a box of books $15\ \text{m}$ across the horizontal floor of the gymnasium by pushing with a horizontal force of $120\ \text{N}$, while kinetic friction opposes the motion with $40\ \text{N}$, as shown in the figure.
- a) Calculate the work done by the student.
- b) Calculate the work done by friction (remember it is negative: $\alpha = 180^\circ$).
- c) Calculate the work done by weight and by the normal force, justifying their values.
- d) Find the net work done on the box.
A musher pulls their team's sled using a rope at an angle of $37^\circ$ to the horizontal, exerting a constant force of $100\ \text{N}$ over $50\ \text{m}$ (…
A musher pulls their team's sled using a rope at an angle of $37^\circ$ to the horizontal, exerting a constant force of $100\ \text{N}$ over $50\ \text{m}$ ($\sen 37^\circ \approx 0.6$; $\cos 37^\circ \approx 0.8$). Friction with the snow opposes with $30\ \text{N}$.
- a) Calculate the work done by the musher.
- b) Calculate the work done by friction.
- c) Find the net work done on the sled.
Work in components (dot product)
A probe's thrusters apply $\vec F = (12, -5)\ \text{N}$ while it moves from $(0, 0)$ to $(8, 3)\ \text{m}$. Measuring the angle between vectors is not practical. How do we calculate the transferred energy using only coordinates?
The displacement vector
If you know the initial point $A(x_A, y_A)$ and the final point $B(x_B, y_B)$, the displacement is the final position minus the initial position: $\vec d = \vec{AB} = (x_B - x_A,\ y_B - y_A)$. Notice the order: final minus initial.
In the laboratory, a trolley on an air track moves $\vec{d} = (5, 0)\ \text{m}$ while acted upon by a constant force $\vec{F} = (6, 8)\ \text{N}$, as shown in the…
In the laboratory, a trolley on an air track moves $\vec{d} = (5, 0)\ \text{m}$ while acted upon by a constant force $\vec{F} = (6, 8)\ \text{N}$, as shown in the figure.
- a) Calculate the work using the dot product $W = \vec{F} \cdot \vec{d} = F_x \cdot d_x + F_y \cdot d_y$.
- b) Explain why the vertical component of the force does not contribute to the work.
A constant force $\vec{F} = (15, -8)\ \text{N}$ acts on a machine component while the component moves through a displacement of $\vec{d} = (4, 2)\ \text{m}$, as shown in…
A constant force $\vec{F} = (15, -8)\ \text{N}$ acts on a machine component while the component moves through a displacement of $\vec{d} = (4, 2)\ \text{m}$, as shown in the figure.
- a) Calculate the work using the dot product $W = \vec{F} \cdot \vec{d} = F_x \cdot d_x + F_y \cdot d_y$.
- b) Check that the component $F_y$ does negative work and interpret what that sign means physically.
On a laboratory air table, a puck of $4\ \text{kg}$, initially at rest at point $A(1, 2)$, moves to point $B(9, 5)$ while a constant force $\vec{F} = (6, 8)\ \text{N}$…
On a laboratory air table, a puck of $4\ \text{kg}$, initially at rest at point $A(1, 2)$, moves to point $B(9, 5)$ while a constant force $\vec{F} = (6, 8)\ \text{N}$ acts on it (the coordinates are in meters), as shown in the figure.
- a) Determine the displacement vector $\vec{d} = \vec{AB}$.
- b) Calculate the work done by the force using the scalar product $W = \vec{F} \cdot \vec{d}$.
- c) Since friction is negligible, this work is the net work: apply the work-energy theorem and calculate the speed of the puck when it reaches $B$.
Weight, ramps, and the inclined plane
When you lift an object, the weight points downward and the displacement upward: $\alpha = 180^\circ$ and the work done by weight is negative. Whoever lifts the object does positive work against the weight.
On a ramp, only the component of the weight parallel to the surface, $P\sin\theta$, does work. The normal force is always perpendicular to the surface, so its work is zero. If there is friction, remember that $N$ is no longer $m\,g$: on the ramp $N = m\,g\cos\theta$.
You place a toolbox of $10\ \text{kg}$ onto a shelf located $2.5\ \text{m}$ above the floor (take $g = 9.8\ \text{m/s}^2$). a) Calculate the weight of the box. b)…
You place a toolbox of $10\ \text{kg}$ onto a shelf located $2.5\ \text{m}$ above the floor (take $g = 9.8\ \text{m/s}^2$).
- a) Calculate the weight of the box.
- b) Calculate the work done by the weight while raising it. Remember that weight points downwards and displacement upwards.
- c) What is the minimum work that you do against gravity?
To load a fridge of $60\ \text{kg}$ into a van, it is pushed up a ramp of $3\ \text{m}$ at an angle of $20^\circ$ to the horizontal ($\sen 20^\circ \approx 0.34$; take…
To load a fridge of $60\ \text{kg}$ into a van, it is pushed up a ramp of $3\ \text{m}$ at an angle of $20^\circ$ to the horizontal ($\sen 20^\circ \approx 0.34$; take $g = 10\ \text{m/s}^2$), as shown in the figure.
- a) Calculate the component of the weight parallel to the ramp.
- b) Calculate the work done by the weight during the ascent.
- c) What minimum work must the person pushing do against that component of weight?
A block of $10\ \text{kg}$ is pulled up a frictionless inclined plane of $30^\circ$ using a force $\vec{F}$ parallel to the plane, so that the block ascends at a…
A block of $10\ \text{kg}$ is pulled up a frictionless inclined plane of $30^\circ$ using a force $\vec{F}$ parallel to the plane, so that the block ascends at a constant speed along $5\ \text{m}$ ($\sen 30^\circ = 0.5$; take $g = 10\ \text{m/s}^2$), as shown in the figure.
- a) Calculate the value of $F$ for the speed to be constant.
- b) Calculate the work done by each force: the applied force, the weight, and the normal force.
- c) Find the net work and explain, using the work-energy theorem, why that result was to be expected.
A worker pulls a $25\ \text{kg}$ box up a $20\ \text{m}$ ramp inclined at $37^\circ$ to the horizontal, pulling with a rope parallel to the ramp and at constant speed…
A worker pulls a $25\ \text{kg}$ box up a $20\ \text{m}$ ramp inclined at $37^\circ$ to the horizontal, pulling with a rope parallel to the ramp and at constant speed ($\sen 37^\circ \approx 0.6$; $\cos 37^\circ \approx 0.8$; coefficient of friction $\mu = 0.25$; take $g = 10\ \text{m/s}^2$), as shown in the figure.
- a) Calculate the normal force exerted by the ramp and the friction force.
- b) Calculate the force with which the worker pulls.
- c) Calculate the work done by each force (pull, weight, friction, and normal force) and check that the net work is zero. Why did it have to be zero?
The work–kinetic energy theorem
Road safety tests show that a car braking hard at $40\ \text{km/h}$ travels about $8\ \text{m}$, but at $80\ \text{km/h}$ it doesn't travel double, but about $32\ \text{m}$. What principle connects the work done by the brakes with the loss of speed without calculating the time?
It is the precise way of saying that energy is transferred by work: if the forces deliver net energy to the body, its kinetic energy increases; if they take it away, it decreases.
The work-energy theorem states that the net work done on a body is equal to the change in its kinetic energy:…
The work-energy theorem states that the net work done on a body is equal to the change in its kinetic energy: $W_{neto} = \Delta E_c = \frac{1}{2} \times m \times v^2 - \frac{1}{2} \times m \times v_0^2$. A demonstration ball of $2\ \text{kg}$ increases its speed from $3\ \text{m/s}$ to $5\ \text{m/s}$ on an air track.
- a) Calculate its initial kinetic energy and its final kinetic energy.
- b) What is the net work done on the ball?
A passenger car of $1200\ \text{kg}$ accelerates from rest to $108\ \text{km/h}$ when joining a dual carriageway. a) Express the final velocity in $\text{m/s}$. b)…
A passenger car of $1200\ \text{kg}$ accelerates from rest to $108\ \text{km/h}$ when joining a dual carriageway.
- a) Express the final velocity in $\text{m/s}$.
- b) Calculate the change in kinetic energy of the car.
- c) What net work has been done by the forces acting on the car? Express it in $\text{kJ}$.
A car of $1000\ \text{kg}$ travelling at $20\ \text{m/s}$ brakes to a halt over a distance of $50\ \text{m}$ along a horizontal section. a) Calculate the change in…
A car of $1000\ \text{kg}$ travelling at $20\ \text{m/s}$ brakes to a halt over a distance of $50\ \text{m}$ along a horizontal section.
- a) Calculate the change in kinetic energy during braking.
- b) Applying the work-energy theorem, find the work done by the net braking force.
- c) What is the average value of this net braking force? State its sense.
A box of $25\ \text{kg}$ starts from rest on the horizontal floor of a warehouse, where the coefficient of kinetic friction is $\mu = 0.2$. It is pushed with a constant…
A box of $25\ \text{kg}$ starts from rest on the horizontal floor of a warehouse, where the coefficient of kinetic friction is $\mu = 0.2$. It is pushed with a constant horizontal force of $100\ \text{N}$ over $12\ \text{m}$ (take $g = 10\ \text{m/s}^2$), as shown in the figure.
- a) Calculate the kinetic friction force, $F_{roz} = \mu \times N$, taking into account that on a horizontal floor the normal force is $N = m \times g$.
- b) Calculate the work done by each of the forces (push, friction, weight, and normal force) and the net work.
- c) Apply the work-energy theorem to find the speed of the box at the end of the $12\ \text{m}$.
A passenger car of $1000\ \text{kg}$ travels at $20\ \text{m/s}$ along a horizontal road and locks its wheels during emergency braking. The coefficient of friction…
A passenger car of $1000\ \text{kg}$ travels at $20\ \text{m/s}$ along a horizontal road and locks its wheels during emergency braking. The coefficient of friction between the tyres and the asphalt is $\mu = 0.5$ (take $g = 10\ \text{m/s}^2$).
- a) Calculate the friction force during braking.
- b) Apply the work-energy theorem to calculate the distance it travels until coming to a stop.
- c) Reason, from the expression obtained, what happens to the braking distance if the initial speed is doubled: does it also double?
On an ice rink, a puck of mass $m$ is launched with an initial speed of $v_0 = 10\ \text{m/s}$, and slides in a straight line until coming to a stop due to friction…
On an ice rink, a puck of mass $m$ is launched with an initial speed of $v_0 = 10\ \text{m/s}$, and slides in a straight line until coming to a stop due to friction ($\mu = 0.25$; take $g = 10\ \text{m/s}^2$).
- a) Write the expression for the work done by friction and, applying the work-energy theorem, obtain the stopping distance $d$ as a function of $v_0$, $\mu$, and $g$.
- b) Calculate this distance numerically.
- c) Reason, from the expression obtained, whether the stopping distance depends on the mass of the puck. What practical implication does this result have for road safety?
An elevator cabin with its passengers has a mass of $600\ \text{kg}$. It starts from rest and ascends $12\ \text{m}$ with a constant acceleration of $1.2\ \text{m/s}^2$…
An elevator cabin with its passengers has a mass of $600\ \text{kg}$. It starts from rest and ascends $12\ \text{m}$ with a constant acceleration of $1.2\ \text{m/s}^2$ (take $g = 10\ \text{m/s}^2$).
- a) Calculate the force exerted by the cable applying Newton's second law: $T - m \times g = m \times a$.
- b) Calculate the work done by the cable and the work done by weight during these $12\ \text{m}$.
- c) Find the net work and, using the work-energy theorem, the speed of the cabin after $12\ \text{m}$. Check the result using kinematics: $v^2 = 2 \times a \times h$.
The work done by weight does not depend on the path
If zero net work means constant speed (the kinetic energy does not change), a crane lowering a beam at constant speed gives us a clear case: the weight does positive work and the cable does the same amount of work, but negative.
A construction crane lowers a beam of $500\ \text{kg}$ from a height of $8\ \text{m}$ to the ground at constant speed (take $g = 9.8\ \text{m/s}^2$). a) Calculate the…
A construction crane lowers a beam of $500\ \text{kg}$ from a height of $8\ \text{m}$ to the ground at constant speed (take $g = 9.8\ \text{m/s}^2$).
- a) Calculate the work done by the weight during the descent.
- b) Calculate the work done by the tension in the cable.
- c) What is the net work? Relate this to the fact that the speed is constant, using the work-energy theorem.
A firefighter of $80\ \text{kg}$ with their equipment climbs a fire escape from the ground to a terrace located $18\ \text{m}$ higher up (take $g = 10\ \text{m/s}^2$).…
A firefighter of $80\ \text{kg}$ with their equipment climbs a fire escape from the ground to a terrace located $18\ \text{m}$ higher up (take $g = 10\ \text{m/s}^2$).
- a) Calculate the work done by the weight during the ascent.
- b) Calculate the work done by the firefighter against gravity.
- c) If instead of the fire escape they had climbed a longer ramp to the same terrace, how would that work change? Justify your answer.
A cable car cabin with its passengers has a total mass of $800\ \text{kg}$. On the outbound journey, it ascends from the base station, at an altitude of…
A cable car cabin with its passengers has a total mass of $800\ \text{kg}$. On the outbound journey, it ascends from the base station, at an altitude of $1200\ \text{m}$, to the top station, at $1750\ \text{m}$. It then partially descends to an intermediate station located at $1550\ \text{m}$ (take $g = 10\ \text{m/s}^2$), as shown in the figure.
- a) Calculate the work done by weight during the ascent.
- b) Calculate the work done by weight during the partial descent.
- c) Calculate the total work done by weight over the entire journey and verify that it depends only on the net difference in altitude between the starting and ending points. What property of the gravitational force does this result highlight?
Back to the beginning: for the pallet pushed in a straight line, all the force is parallel to the displacement ($\alpha = 0^\circ$, maximum transfer); for the cable car cabin, the work done by weight depends only on the net height difference; and in every case the net work is the energy that enters or leaves the body as kinetic energy.