Work and Energy

Work and Energy


In this lesson, we will learn:

  • Work done on an object is the change in an object's mechanical energy
  • Work done by a force is the product of the force and the displacement of the object


  • Work is the transfer of energy from one place to another. Because work is a form of energy, it is a scalar and measured in joules (J).
  • Work is done when a force moves an object over a displacement. It is equal to force times displacement, or W=FdW = F_\parallel d. It can be either positive or negative. Doing positive work on an object increases the object's mechanical energy. Positive work can increase an object's kinetic energy (by accelerating it), potential energy (by moving it to a greater height), or both. This can be expressed with the equation W=Fd=ΔEmechW = F_\parallel d = \Delta E_{mech}. Negative work on an object reduces its mechanical energy.
    • The parallel sign ("\parallel") in the formula W=FdW = F_\parallel d indicates that only a force that is parallel to the displacement of an object can do work. Remember, in order for a force to do positive work on an object (add energy), it has to add kinetic and/or potential energy to the object. Consider the following examples:
      • A force pointed in the same direction as an object's displacement does positive work.
      • A force that does not cause a displacement does not do work.
      • A force pointed perpendicular to an object's displacement does not work.
      • If a force is applied to an object and the resulting displacement is at a non-90° angle from the force (i.e. moving an object by pushing/pulling at an angle), only the component of the force that is pointed in the same direction as the displacement does work. The other component is perpendicular to the displacement and does no work.
      • A force that points in the opposite direction of an object's displacement does negative work.
        • A common force that does negative work is friction, since it is always pointed opposite the direction of motion.


      W=Fd=ΔEmech=(Ekf+Epf)(Eki+Epi)W = F_\parallel d = \Delta E_{mech} = (E_{kf} + E_{pf}) - (E_{ki} + E_{pi})

      W:W: work, in joules (J)

      d:d: displacement, in meters (m)

      F:F_\parallel: component of force parallel to dd in newtons (N)

      ΔEmech:\Delta E_{mech}: change in mechanical energy

      (Ekf+Epf):(E_{kf} + E_{pf}): total final mechanical energy, in joules (J)

      (Eki+Epi):(E_{ki} + E_{pi}): total initial of force parallel to dd, in newtons (N)

      Kinetic Energy

      Ek=12mv2E_k = \frac{1}{2}mv^2

      Ek;E_k; kinetic energy, in joules (J)

      m:m: mass, in kilgrams (kg)

      v:v: velocity, in meters per second (m/s)

      Gravitational Potential Energy

      Ep=mghE_p = mgh

      Ep:E_p: gravitational potential energy, in joules (J)

      g:g: acceleration due to gravity, in meters per second squared (m/s2)

      h:h: height, in meters (m)

  • 1.
    W=ΔEmech=Fd:\bold{W = \Delta E_{mech} = F_\parallel d:} Calculating work
    A 1170 kg car travels at 11.0 m/s.
    1. How much work needs to be done on the car to accelerate it to 24.0 m/s?
    2. What is the net force acting on the car that accelerates it, if the acceleration is uniform and happens over 95.0 m?

    A 5.50 kg box slides across a floor at 12.0 m/s. Friction slows the box to 2.00 m/s after is has travelled 13.0 m. Find the work done on the box and the force of friction acting on the box. 

    A 2.50 kg box is initially at rest at the top of a 30.0° slope and reaches a speed of 11.5 m/s when it slides 12.0 m down the slope. Find the force of friction acting on the box.

  • 2.
    W=ΔEmech=Fd:\bold{W = \Delta E_{mech} = F_\parallel d:} Calculating work with force applied at an angle
    A force of 885 N pulls on a box at an angle of 28.0° above the horizontal. 115 N of friction acts on the box as it slides 19.0 m. How much work does the 885 N force do on the box?