Solved Problem on Work and Energy
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A weight of 10 N is lifted from the rest, by a force of 30 N to a height of 5 m. Determine:
a) The work done by the force applied to raise the weight to the height given;
b) The work done by the gravitational force;
c) The speed of the body while reaching the height given.


Problem data:
  • Weight of body:    W = 10 N;
  • Force applied to the body:    F = 30 N;
  • Height as far as the body is lifted:    h = 5 m;
  • Initial speed of body:    v0 = 0;
  • Acceleration due to gravity:    g = 9.8 m/s2.
Problem diagram:

We choose a frame of reference on the floor and oriented upward, in this case, the acceleration due to gravity and the gravitational force will be negative (Figure 1).
Figure 1

Solution

a) The work done by the force is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {\mathscr{W}_{\small F}=Fd} \tag{I} \end{gather} \]
in this case, the force is the external force, F, that raises the load, and the distance will be the height, h, to which the load was lifted
\[ \begin{gather} \mathscr{W}_{\small F}=Fh\\[5pt] \mathscr{W}_{\small F}=(30\;\mathrm N)\times(5\;\mathrm m) \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {\mathscr{W}_{\small F}=150\;\mathrm J} \end{gather} \]

b) Applying the equation (I) to the body
\[ \begin{gather} \mathscr{W}_{\small W}=Wh\\[5pt] \mathscr{W}_{\small W}=(-10\;\mathrm N)\times(5\;\mathrm N) \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {\mathscr{W}_{\small W}=-50\;\mathrm{J}} \end{gather} \]

Note: there is no negative energy, the negative sign indicates that the work is done in the opposite direction to the force due to gravity. The force due to gravity points to Earth and the object is pulled away when lifted


c) The resultant force, FR, between external force and gravitational force is equal to
\[ \begin{gather} F_{\small R}=F-W\\[5pt] F_{\small R}=30\;\mathrm N-10\;\mathrm N\\[5pt] F_{\small R}=20\;\mathrm N \end{gather} \]
the gravitational force is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {F_g=mg} \end{gather} \]
\[ \begin{gather} m=\frac{F_g}{g}\\[5pt] m=\frac{-10\;\mathrm N}{-9.8\;\mathrm{\frac{m}{s^2}}}\\[5pt] m\approx1\;\mathrm{kg} \end{gather} \]
According to the Work-Energy Kinetic Theorem, the work of a force is equal to the change of Kinetic Energy
\[ \begin{gather} \bbox[#99CCFF,10px] {\mathscr{W}_{\small F}=\Delta K=K_f-K_i} \tag{II} \end{gather} \]
the kinetic energy is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {K=\frac{mv^2}{2}} \tag{III} \end{gather} \]
substituting equations (I) and (III) into equation (II) the work done by the resultant force to lift the body is given by
\[ \begin{gather} F_{\small R}h=\frac{mv^2}{2}-\frac{mv_0^2}{2}\\[5pt] v=\sqrt{\frac{2}{m}\left[F_{\small R}h+\frac{mv_0^2}{2}\right]\;}\\[5pt] v=\sqrt{\frac{2}{1\;\mathrm{kg}}\left[(20\;\mathrm N)\times(5\;\mathrm m)+\frac{(1\;\mathrm{kg})\times(0)^2}{2}\right]\;}\\[5pt] v=\sqrt{\frac{2}{1\;\mathrm{\cancel{kg}}}\times\left[100\;\mathrm{\frac{\cancel{kg}.m^2}{s^2}}\right]\;} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {v\approx 14.1\;\mathrm{m/s}} \end{gather} \]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .