Courses for Kids
Free study material
Offline Centres
Store Icon

In the given figure a $0.250{\text{kg}}$ block of cheese lies on the floor of a $900{\text{kg}}$ elevator cab that is being pulled upward by a cable through a distance ${d_1} = 2.40\;{\text{m}}$ and then through a distance ${d_2} = 10.5m$.
$(a)$ Through ${d_1}$, if the normal force on the block from the floor has constant magnitude ${F_N} = 3.00\;{\text{N}}$, how much work is done on the cab by the force from the cable?
$(b)$ Through ${d_2}$, if the work done on the cab by the (constant) force from the cable is $92.61\;{\text{kJ}}$, what is the magnitude of ${F_N}$?

Last updated date: 13th Jun 2024
Total views: 51.9k
Views today: 0.51k
51.9k+ views
Hint: In this question, the concept of the force equilibrium will be used. First just do force balances, one for the block of cheese and one for the elevator. From this we can get the net force pulling the elevator upward, and from that and the distance covered we can find the work.

Complete step by step solution:
For the cheese, it has a normal force upward and its weight downward, and this must equal its mass times acceleration. Its acceleration is upward, which we will choose as the positive direction. This makes the normal force positive and the weight negative.
Cheese alone:
\[{F_N} - {m_c}g = {m_c}a\]
\[ \Rightarrow a = \dfrac{{{F_N}{\text{ }}}}{{{m_c}}} - g\]
Now, we substitute the given values in the above expression as,
\[ \Rightarrow a = \dfrac{3}{{0.25}} - 9.81\]
\[ \Rightarrow a \approx 2.19\;{\text{m/}}{{\text{s}}^{\text{2}}}{\text{ }}\]
This gives us the acceleration of the cheese. Balancing forces for the elevator as a whole get us the net force \[F\]. On the elevator, we have \[F\] acting upward and its weight pulling downward, giving an overall positive acceleration \[a\] which is the same as the cheese. Note that the weight here is that of the elevator plus the cheese, though the latter is utterly negligible in this force balance.
Elevator plus cheese:
\[F - \left( {{m_e}{\text{ }} + {\text{ }}{m_c}} \right){\text{ }}g = \left( {{m_e} + {m_c}} \right)a\]
\[ \Rightarrow F = \left( {{m_e}{\text{ }} + {\text{ }}{m_c}} \right)\left( {g{\text{ }} + {\text{ }}a} \right)\]
\[ \Rightarrow F = \left( {{m_e}{\text{ }} + {\text{ }}{m_c}} \right)\left( {g{\text{ }} + {\text{ }}\dfrac{{{F_n}{\text{ }}}}{{{m_c}}}{\text{ }} - {\text{ }}g{\text{ }}} \right)\]
Now, we simplify further to get,
\[ \Rightarrow F = {\text{ }}\left( {\dfrac{{{m_e} + {m_c}}}{{{m_c}}}} \right){F_n}\]
The work done by the external force is the force \[F\] is this force times the displacement. For the first part of the motion, the displacement is\[{d_1}\].
\[W = F{\text{ }}{d_1}\]
\[ \Rightarrow W = \left( {\dfrac{{{m_e}{\text{ }} + {\text{ }}{m_c}}}{{{m_c}}}} \right){F_n}{d_1}\]
Now, we substitute the given values in the above expression as,
   \Rightarrow W = \left( {\dfrac{{{\text{900}} + 0.25}}{{0.25}}} \right)\left( 3 \right)\left( {2.4} \right){\text{ }} \\
  \therefore W \approx {\text{ }}25.9{\text{ kJ}} \\

Thus, the answer to part \[(a)\] is \[25.9{\text{kJ}}\].

For the second part, given the work done \[W\], displacement \[{d_2}\], and masses, we can solve for the normal force as,
\[{F_{n{\text{ }}}} = \dfrac{{W{m_c}}}{{\left( {{m_e}{\text{ }} + {\text{ }}{m_c}} \right){d_2}}}\]
Now, we substitute the given values in the above expression as,
\[ \Rightarrow {F_{n{\text{ }}}} = \dfrac{{\left( {92.61} \right)\left( {0.25} \right)}}{{\left( {900 + 0.25} \right)\left( {10.5} \right)}}\]
\[\therefore {F_{n{\text{ }}}} \approx 2.45\;{\text{N}}\]

Thus, the answer to part \[(b)\]is \[2.45{\text{N}}\].

Note: As we know that the force is the vector quantity that has magnitude as well as direction, but the work done is the scalar quantity which has only magnitude instead of direction. So, the work done is the dot product of the force and displacement.