A traveling salesman is trying to sell an umbrella. He claims that its novel spring mechanism opens up the umbrella in just 0.5 second and “uses only 1 kilo-watt of power” to open. He also shows off the slender 30 cm long bag that holds the umbrella when it is not in use.

a) Assuming the traveling salesman is to be believed about the power of the umbrella, calculate the change in the umbrella’s mechanical energy that is transferred from the chemical potential energy stored in the person’s muscles while opening the umbrella.

b) Assuming the spring inside the umbrella is at equilibrium when the umbrella is open (and is 1 meter in length), find the elastic spring constant.

c) To make the umbrella appear more impressive, the vendor claims that the energy stored in the compressed umbrella is equivalent to the same energy needed to lift a 500 kg object to a 1 meter height. Is this claim true, or is this a case of false advertising?

a) Write the equation for power in terms of energy, and solving for the answer should then become fairly straightforward.

b) Relate the energy to the elastic potential energy, and you can then solve for the spring constant.

c) Use the formula for the gravitational potential energy, and plug in the numbers.

a) Power is defined as energy over time. We can then write:

\begin{equation*}
P\approx \frac{\Delta E}{\Delta t}.
\end{equation*}

Solving for \(\Delta E \), and plugging in numerical values, we get:

\begin{equation*}
\Delta E = 500 \, \text{J}.
\end{equation*}

b) The elastic potential energy is given as:

\begin{equation*}
\Delta E = \frac{1}{2} kx^2.
\end{equation*}

The deformation is \(x=0.7 \, \text{m}\). Solving for \(k\), we get:

\begin{equation*}
k \approx 2041 \, \text{N/m}.
\end{equation*}

c) The gravitational potential energy is:

\begin{equation*}
U_g = mg\Delta h.
\end{equation*}

which, with numerical values, is:

\begin{equation*}
U_g = 4900 \, \text{J}.
\end{equation*}

a) The energy released when the umbrella opens can be estimated using the power \(P=1\text{kW}=1000\,\text{W}\) and the time required to open the umbrella \(\Delta t=0.5\,\text{s}\). If the time \(\Delta t\) is small we can approximate the power as

\begin{equation}
P\approx \frac{\Delta E}{\Delta t},
\end{equation}

where \(\Delta E\) is the amount of energy released when the umbrella opens. Solving for \(\Delta E\) from the equation above we get

\begin{equation}
\Delta E=P\Delta t.
\end{equation}

Using the numerical values we have

\begin{equation}
\Delta E=(1000\,\text{W})(0.5\,\text{s}),
\end{equation}

\begin{equation}
\label{delta}
\Delta E=500\,\text{J}.
\end{equation}

b) We need to find now the elastic constant of the spring inside the umbrella. Recall that the energy released by the umbrella as it opens is due to the potential elastic energy stored in the spring inside. This means that the energy stored in the spring mechanism when the umbrella is closed (the elastic potential energy) is the same as the energy \(\Delta E\) released by the umbrella when it is opened:

\begin{equation}
\label{de}
\Delta E=\frac{1}{2}kx^2,
\end{equation}

where \(k\) is the spring’s elastic constant and \(x\) the deformation of the spring with respect to its equilibrium position. We know that the open umbrella measures \(1\,\text{m}\) and that when it is closed it fits in a \(30\,\text{cm}=0.3\,\text{m}\) bag. Then the deformation is

\begin{equation}
x=1\,\text{m}-0.3\,\text{m}=0.7\,\text{m}.
\end{equation}

Solving for \(k\) in equation \eqref{de} we obtain

\begin{equation}
k=\frac{2\Delta E}{x^2},
\end{equation}

which, after using the numerical values, yields

\begin{equation}
k=\frac{2(500\,\text{J})}{(0.7\,\text{m})^2},
\end{equation}

\begin{equation}
k\approx 2041\,\text{N/m}.
\end{equation}

c) The energy necessary to lift a mass of \(m=500\,\text{kg}\) an amount \(h=1\,\text{m}\) is given by the change in potential gravitational energy \(U_g\), namely

\begin{equation}
U_g=mg\Delta h,
\end{equation}

where \(g=9.8\,\text{m/s}^2\) is the gravitational acceleration on Earth and \(\Delta h\) is the change in height. Using the numerical values, we get for the amount of energy

\begin{equation}
U_g=(500\,\text{kg})(9.8\,\text{m/s}^2)(1\,\text{m}),
\end{equation}

\begin{equation}
U_g=4900\,\text{J}.
\end{equation}

Comparing the value of \(U_g\) with that of \(\Delta E\) in equation \eqref{delta}, we see that \(U_g\) is larger by an order of magnitude. Thus, the vendor’s claim is not correct.

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