Question

A worker eats a dish of ice cream which furnishes 200 kcal of heat. If the efficiency of the body is 25% then calculate the height to which the worker can carry a weight of 56 kg using the energy of ice cream. Worker’s own weight is 70 kg. Given \[1\,{\text{cal}} = 4.2\,{\text{J}}\]
A. 145 m
B. 170.07 m
C. 128 m
D. 135 m

Answer 1

Hint: The energy furnished by the ice cream will be spent to overcome the potential energy required to raise the weight. Determine the heat energy provided by the ice cream in joule. The potential energy at height h will be equal to 25% of the energy provided by the ice cream.

Formula used:
Potential energy, \[U = mgh\]
where, m is the mass, g is the acceleration due to gravity and h is the height.

Complete step by step answer:
As the dish of ice cream furnishes the energy to the worker, this energy is used by the worker to raise the weight of 56 kg to a certain height h. The energy furnished by the ice cream will be spent to overcome the potential energy required to raise the weight.

We have given the energy furnished by the ice cream,
\[Q = 200\,{\text{kcal}} = 200 \times {10^3}\,{\text{cal}}\]
\[ \Rightarrow Q = \left( {200 \times {{10}^3}\,{\text{cal}}} \right)\left( {\dfrac{{4.2\,{\text{J}}}}{{{\text{1}}\,{\text{cal}}}}} \right)\]
\[ \Rightarrow Q = 8.4 \times {10^5}\,{\text{J}}\]
We have given that only 25% of this energy is used to cover the height h. Therefore, the energy used by the worker is,
\[25\% Q = \dfrac{{8.4 \times {{10}^5}\,{\text{J}}}}{4}\]
\[ \Rightarrow Q = 2.1 \times {10^5}\,{\text{J}}\]

Now, this much energy is used to overcome the potential energy. Therefore, we can write,
\[Q = \left( {M + m} \right)gh\]
Here, M is the mass of the worker, m is the mass of the weight, g is the acceleration due to gravity and h is the height to which the weight can be raised by the worker.
Substituting \[2.1 \times {10^5}\,{\text{J}}\] for Q, 70 kg for M, 56 kg for m and \[9.8\,{\text{m/}}{{\text{s}}^2}\] for g in the above equation, we get,
\[2.1 \times {10^5} = \left( {70 + 56} \right)\left( {9.8} \right)h\]
\[ \Rightarrow h = \dfrac{{2.1 \times {{10}^5}}}{{1234.8}}\]
\[ \therefore h = 170.07\,{\text{m}}\]

So, the correct answer is option B.

Note:Do not consider the mass of only the weight that is being carried by the worker. The mass of the worker is also included in the potential energy term. The energy provided by the food is spent to rise to the height due to increase in the potential energy. Note that the S.I unit of energy is joule and not calorie.