Worker Dragging a Crate

Task number: 1983

A worker drags a crate across a smooth horizontal floor. The worker exerts a force of magnitude 210 N on the rope, which is inclined at an upward angle 20° to the horizontal. Determine the work performed by the following forces provided that the crate was moved by 0.5 m:

a) force exerted by the worker

b) gravitational force

c) normal force exerted by the floor on the crate.

d) What is the total work performed by all of the forces above?

Neglect the friction of the crate on the floor.

  • Given values

    m = 50 kg Mass of the crate
    F = 210 N Magnitude of the force exerted by the worker on the crate
    α = 20° Angle formed by the force vector \(\vec{F}\) and the displacement vector
    s = 0.5 m Displacement
    W = ? Work performed by force \(\vec{F}\)
    Wg = ? Work performed by the gravitational force
    WN = ? Work performed by the normal force
    WC = ? Total work
  • Hint 1 – Work performed by the worker

    Draw a picture and mark all of the forces acting on the crate.

    The force exerted by the worker can be considered constant. What is the relation for work performed by a constant force?

  • Hint 2 – Work performed by gravitational force

    In the relation (1) for a constant force substitute the force \(\vec{F}\) by the relation for gravitational force. What is the angle between the gravitational force and the displacement vector?

  • Hint 3 – Work performed by normal Force

    In the relation (1) substitute the force \(\vec{F}\) by the normal force \(\vec{N}\).

  • Hint 4 – Total work

    The total work can be determined by the sum of works performed by individual forces.

  • Numerical solution

    a)

    \[F\,=\, 210\, \mathrm{N}\] \[s\,=\, 0.5 \,\mathrm{m}\] \[\alpha\,=\,20^{\circ}\]
    \[W\,=\,Fs\cos\alpha\] \[W\,=\,210{\cdot}0.5\cdot\cos20^{\circ}\,\mathrm{J}\] \[W\,=\,105{\cdot}0.94\,\mathrm{J}\] \[W\,=\,98.7\,\mathrm{J}\dot=\, 99\,\mathrm{J}\]

    b)

    The gravitational force vector forms an angle of 90° with the displacement vector. Therefore the work performed by this force is zero.

    c)

    The normal force vector forms an angle of 90° with the displacement vector. Therefore the work performed by this force is zero.

    d)

    \[W\,=\,98.7\,\mathrm{J}\] \[W_\mathrm{g}\,=\,0\,\mathrm{J}\] \[W_\mathrm{N}\,=\,0\,\mathrm{J}\] \[W_\mathrm{c}\,=\,W+W_\mathrm{g}+W_\mathrm{N}\] \[W_\mathrm{c}\,=\,98.7\,\mathrm{J} \dot=\, 99\,\mathrm{J}\]
  • Answer

    a) The work performed by the worker is \(W\,=\,Fs\cos\alpha\,\dot=\, 99\,\mathrm{J}\).

    b) The work performed by the gravitational force is zero.

    c) The work performed by the normal force is zero.

    d) The total work is \(W_\mathrm{c}\,=\,W+W_\mathrm{g}+W_\mathrm{N}\,=\,Fs\cos\alpha\,\dot=\, 99\,\mathrm{J}\).

  • Comment

    The motion of the crate is a uniformly accelerated motion.

    In this task we solve an idealized problem, in which we do not take into account the frictional force, which would in a real situation act in the direction opposite to the force \(\vec{F}\).

  • Overall solution of a) – Work performed by the worker

    Worker dragging a crate

    \(\vec{F}\) … force exerted by the worker on the crate

    \(\vec{N}\) … normal force exerted by the floor

    \(\vec{F}_\mathrm{g}\) … gravitational force

    If the force acting on the body is constant, we determine the work performed by this force to be a product of the force vector and the displacement vector.

    \[W\,=\,\vec{F}\cdot\vec{s}\]

    \(W\) … performed work

    \(\vec{F}\) … force vector exerted by the worker

    \(\vec{s}\) … displacement vector

    For the scalar product it holds:

    \[W\,=\,Fs\cos \alpha.\tag{1}\] \[F= 210 \,\mathrm{N},\] \[s= 0{,}5\,\mathrm{m},\] \[\alpha=20^{\circ},\] \[W\,=\,Fs\cos\alpha\,=\,210{\cdot}0.5\cdot\cos20^{\circ}\,\mathrm{J}\,=\,105{\cdot}0.94\,\mathrm{J1},\] \[W\,=\,98.7\,\mathrm{J}\dot=\, 99\,\mathrm{J}.\]

    Answer: The work performed by the worker is \[W\,=\,Fs\cos\alpha\,\dot=\, 99\,\mathrm{J}.\]

  • Overall solution of b) – Work performed by the gravitational force

    In the relation (1) we substitute the constant force \(\vec{F}\) by the relation for gravitational force \(\vec{F}_\mathrm{g}\).

    \[W_\mathrm{g}\,=\,\vec{F}_\mathrm{g}\cdot\vec {s}\] \[W_\mathrm{g}\,=\,m\cdot\vec{g}\cdot\vec{s}\] \[W_\mathrm{g}\,=\,mgs\cos \alpha^{'}\]

    \(W_\mathrm{g}\) … work performed by the gravitational force

    \(\vec{F}_\mathrm{g}\) … gravitational force

    \(m\) … mass of the crate

    \(\vec{g}\) … gravitational acceleration

    \(\vec{s}\) … displacement vector

    \(\alpha^{'}\) … angle formed by the vector of the gravitational force and the displacement vector, \(\alpha^{'}=90^{\circ}\)

    \[\cos\alpha^{'}\,=\,\cos90^{\circ}\,=\,0\] \[W_\mathrm{g}\,=\,m\cdot g\cdot s\cdot 0\,=\,0\]

    Answer: The work performed by the gravitational force is zero.

  • Overall solution of c) – Work performed by the normal force

    In the relation (1) substitute the force \(\vec{F}\) by the normal force \(\vec{N}\).

    \[W_\mathrm{N}\,=\,\vec{N}\cdot\vec{s}\] \[W_\mathrm{N}\,=\,Ns\cos\alpha^{''}\]

    \(W_\mathrm{N}\) … work performed by the normal force

    \(\vec{N}\) … normal force

    \(\vec{s}\) … displacement vector

    \(\alpha^{''}\) … angle formed by the vector of the normal force and the displacement vector

    The normal force vector forms an angle of 90° with the displacement vector. The work is therefore zero.

    \[\cos\alpha^{''}\,=\,\cos90^{\circ}\,=\,0\] \[W_\mathrm{N}\,=\,m\cdot g\cdot s\cdot 0\,=\,0\]

    Answer: The work performed by the normal force is zero.

  • Overall solution of d) – Total work

    The total work can be determined the sum of works performed by individual forces

    \[W_\mathrm{c}\,=\,W+W_\mathrm{g}+W_\mathrm{N}\]

    \(W_\mathrm{c}\) … total work

    \(W\) … work performed by the force \(\vec{F}\)

    \(W_\mathrm{g}\) … work performed by the gravitational force

    \(W_\mathrm{N}\) … work performed by the normal force

    \[W_\mathrm{c}\,=\,F\cdot s \cos\alpha\,+\,0\,+\,0\] \[W\,=\,Fs\cos\alpha\] \[W_\mathrm{g}\,=\,0\,\mathrm{J}\] \[W_\mathrm{N}\,=\,0\,\mathrm{J}\] \[W_\mathrm{c}\,=\,W+W_\mathrm{g}+W_\mathrm{N}=Fs\cos\alpha\] \[W_\mathrm{c}\,=\,98.7\,\mathrm{J}\dot=\, 99\,\mathrm{J}\]

    Answer: The total work is \(W_\mathrm{c}\,=\,W+W_\mathrm{g}+W_\mathrm{N}\,=\,Fs\cos\alpha\,\dot=\, 99\,\mathrm{J}\).

Difficulty level: Level 2 – Upper secondary level
Tasks requiring routine calculations
Original source: Halliday,D., Resnick,R., Walker,J.: Fyzika. Vysokoškolská učebnice
obecné fyziky. VUTIUM, Brno 2000 - upraveno
Zpracováno v bakalářské práci Jany Šimkové (2008).
×Original source: Halliday,D., Resnick,R., Walker,J.: Fyzika. Vysokoškolská učebnice obecné fyziky. VUTIUM, Brno 2000 - upraveno
Zpracováno v bakalářské práci Jany Šimkové (2008).
Send comment on task by email