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Question: In the given figure a \(0.250{\text{kg}}\) block of cheese lies on the floor of a \(900{\text{kg}}\)...

In the given figure a 0.250kg0.250{\text{kg}} block of cheese lies on the floor of a 900kg900{\text{kg}} elevator cab that is being pulled upward by a cable through a distance d1=2.40  m{d_1} = 2.40\;{\text{m}} and then through a distance d2=10.5m{d_2} = 10.5m.
(a)(a) Through d1{d_1}, if the normal force on the block from the floor has constant magnitude FN=3.00  N{F_N} = 3.00\;{\text{N}}, how much work is done on the cab by the force from the cable?
(b)(b) Through d2{d_2}, if the work done on the cab by the (constant) force from the cable is 92.61  kJ92.61\;{\text{kJ}}, what is the magnitude of FN{F_N}?

Explanation

Solution

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:
FNmcg=mca{F_N} - {m_c}g = {m_c}a
a=FN mcg\Rightarrow a = \dfrac{{{F_N}{\text{ }}}}{{{m_c}}} - g
Now, we substitute the given values in the above expression as,
a=30.259.81\Rightarrow a = \dfrac{3}{{0.25}} - 9.81
a2.19  m/s2 \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 FF. On the elevator, we have FF acting upward and its weight pulling downward, giving an overall positive acceleration aa 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(me + mc) g=(me+mc)aF - \left( {{m_e}{\text{ }} + {\text{ }}{m_c}} \right){\text{ }}g = \left( {{m_e} + {m_c}} \right)a
F=(me + mc)(g + a)\Rightarrow F = \left( {{m_e}{\text{ }} + {\text{ }}{m_c}} \right)\left( {g{\text{ }} + {\text{ }}a} \right)
F=(me + mc)(g + Fn mc  g )\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,
F= (me+mcmc)Fn\Rightarrow F = {\text{ }}\left( {\dfrac{{{m_e} + {m_c}}}{{{m_c}}}} \right){F_n}
The work done by the external force is the force FF is this force times the displacement. For the first part of the motion, the displacement isd1{d_1}.
W=F d1W = F{\text{ }}{d_1}
W=(me + mcmc)Fnd1\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,

W=(900+0.250.25)(3)(2.4)  W 25.9 kJ  \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)(a) is 25.9kJ25.9{\text{kJ}}.

For the second part, given the work done WW, displacement d2{d_2}, and masses, we can solve for the normal force as,
Fn =Wmc(me + mc)d2{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,
Fn =(92.61)(0.25)(900+0.25)(10.5)\Rightarrow {F_{n{\text{ }}}} = \dfrac{{\left( {92.61} \right)\left( {0.25} \right)}}{{\left( {900 + 0.25} \right)\left( {10.5} \right)}}
Fn 2.45  N\therefore {F_{n{\text{ }}}} \approx 2.45\;{\text{N}}

Thus, the answer to part (b)(b)is 2.45N2.45{\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.