Question

In the figure, a 0.25 kg block of cheese lies on the floor of a 970 kg elevator cab that is being pulled upward by a cable through distance d1 = 2.3 m and then through distanced2 = 10.2 m. (a) Through d1, if the normal force on the block from the floor has constant magnitude FN = 3.11 N, how much work is done on the cab by the force from the cable? (b) Through d2, if the work done on the cab by the (constant) force from the cable is 91.98 kJ, what is the magnitude of FN?

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Answer #1
Concepts and reason

The concepts of the free body diagram, Newton’s second law and the work done are used to solve this problem.

Initially, calculate the acceleration of the cheese block drawing its free body diagram and applying Newton’s second law. Calculate the magnitude of the tension in the cable by drawing its free body diagram and applying Newton’s second law.

Calculate the amount of work done by the tension in the cable on the elevator cab using the expression of the work done.

Calculate the magnitude of the new tension in the cable using the expression of the work done. Calculate the new acceleration of the elevator cab drawing its free body diagram while it is at a distance of d2{d_2} and applying Newton’s second law.

Calculate the magnitude of the new FN{F_N} drawing its free body diagram of the cheese block and applying the Newton’s second law

Fundamentals

The expression of Newton’s second law is written as,

Fnet=ma{F_{net}} = ma

Here, Fnet{F_{net}} is the net forces acting on the object, mm is the mass of the object and aa is the acceleration of the object.

The expression for the weight of the object is written as,

W=mgW = mg

Here, WW is the weight of the object and gg is the acceleration due to gravity.

The expression of the work done is written as,

W=FdcosθW = Fd\cos \theta

Here, WW is the work done, FF is the force, dd is the displacement and θ\theta is the angle between the force and the displacement

The sign convention for force:

The forces acting in the upward direction are taken as positive and the forces acting in the downward direction are taken as negative.

(a)

Draw the free body diagram of the cheese block.

W =m2g

In the above diagram, W2{W_2} is the weight of the cheese block, FN{F_N} is the normal force or contact force applied by the elevator’s surface to the cheese block, m2{m_2} is the mass of the cheese block and aa is the acceleration of the block.

Apply Newton’s second law in the vertical direction.

FNm2g=m2a{F_N} - {m_2}g = {m_2}a

Rearrange for aa .

a=FNm2gm2a = \frac{{{F_N} - {m_2}g}}{{{m_2}}}

Substitute 3.11N3.11{\rm{ N}} for FN{F_N} , 0.25kg0.25{\rm{ kg}} for m2{m_2} and 9.81m/s29.81{\rm{ m/}}{{\rm{s}}^2} for gg .

a=3.11N0.25kg(9.81m/s2)0.25kg=2.63m/s2\begin{array}{c}\\a = \frac{{3.11{\rm{ N}} - 0.25{\rm{ kg}}\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)}}{{0.25{\rm{ kg}}}}\\\\ = 2.63{\rm{ m/}}{{\rm{s}}^2}\\\end{array}

Draw the free body diagram of the elevator cab.

W =(m, + m2)g

In the above diagram, m1{m_1} is the mass of the elevator, (m1+m2)\left( {{m_1} + {m_2}} \right) is the total mass of the elevator, WW is the total weight of the elevator and TT is the tension in the cable.

Apply Newton’s second law in the vertical direction.

T(m1+m2)g=(m1+m2)aT - \left( {{m_1} + {m_2}} \right)g = \left( {{m_1} + {m_2}} \right)a

Rearrange for TT .

T=(m1+m2)a+(m1+m2)g=(m1+m2)(a+g)\begin{array}{c}\\T = \left( {{m_1} + {m_2}} \right)a + \left( {{m_1} + {m_2}} \right)g\\\\ = \left( {{m_1} + {m_2}} \right)\left( {a + g} \right)\\\end{array}

Substitute 970kg970{\rm{ kg}} for m1{m_1} , 0.25kg0.25{\rm{ kg}} for m2{m_2} , 2.63m/s22.63{\rm{ m/}}{{\rm{s}}^2} for aa and 9.81m/s29.81{\rm{ m/}}{{\rm{s}}^2} for gg .

T=(970kg+0.25kg)(2.63m/s2+9.81m/s2)=12069.91N\begin{array}{c}\\T = \left( {970{\rm{ kg}} + 0.25{\rm{ kg}}} \right)\left( {2.63{\rm{ m/}}{{\rm{s}}^2} + 9.81{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 12069.91{\rm{ N}}\\\end{array}

The elevator is moving straight upward. So, the angle between the force and its displacement is to be zero.

θ=0\theta = 0^\circ

Here, θ\theta is the angle between the force and its displacement.

Write the expression for the net work done by the cable to displace the elevator cab at a distance of d1{d_1} .

WD1=Td1cosθW{D_1} = T{d_1}\cos \theta

Here, WD1W{D_1} is net work done by the cable to displace the elevator cab at a distance of d1{d_1} .

Substitute 12069.91N12069.91{\rm{ N}} for TT , 2.3m2.3{\rm{ m}} for d1{d_1} and 00^\circ for θ\theta .

WD1=(12069.91N)(2.3m)cos0=27760.79J=27.76kJ\begin{array}{c}\\W{D_1} = \left( {12069.91{\rm{ N}}} \right)\left( {2.3{\rm{ m}}} \right)\cos 0^\circ \\\\ = 27760.79{\rm{ J}}\\\\ = {\rm{27}}{\rm{.76 kJ}}\\\end{array}

(b)

Write the expression for the net work done by the cable to displace the elevator cab at a distance of d2{d_2} .

WD2=Td2cosθW{D_2} = T'{d_2}\cos \theta

Here, WD2W{D_2} is net work done by the cable to displace the elevator cab at a distance of d2{d_2} and TT' is the new tension in the cable.

Rearrange for TT' .

T=WD2d2cosθT' = \frac{{W{D_2}}}{{{d_2}\cos \theta }}

Substitute 91.89kJ91.89{\rm{ kJ}} for WD2W{D_2} , 10.2m10.2{\rm{ m}} for d2{d_2} and 00^\circ for θ\theta .

T=91.89kJ(103J1kJ)(10.2m)cos0=9008.82N\begin{array}{c}\\T' = \frac{{91.89{\rm{ kJ}}\left( {\frac{{{{10}^3}{\rm{ J}}}}{{1{\rm{ kJ}}}}} \right)}}{{\left( {10.2{\rm{ m}}} \right)\cos 0^\circ }}\\\\ = 9008.82{\rm{ N}}\\\end{array}

Again, draw the free body diagram of the elevator cab.

W =(m, + m2)g

Apply Newton’s second law in the vertical direction.

T(m1+m2)g=(m1+m2)aT' - \left( {{m_1} + {m_2}} \right)g = \left( {{m_1} + {m_2}} \right)a'

Rearrange for aa .

a=T(m1+m2)g(m1+m2)a' = \frac{{T' - \left( {{m_1} + {m_2}} \right)g}}{{\left( {{m_1} + {m_2}} \right)}}

Substitute 970kg970{\rm{ kg}} for m1{m_1} , 0.25kg0.25{\rm{ kg}} for m2{m_2} , 9008.82N9008.82{\rm{ N}} for TT' and 9.81m/s29.81{\rm{ m/}}{{\rm{s}}^2} for gg .

a=(9008.82N)(970kg+0.25kg)(9.81m/s2)(970kg+0.25kg)=0.52m/s2\begin{array}{c}\\a' = \frac{{\left( {9008.82{\rm{ N}}} \right) - \left( {970{\rm{ kg}} + 0.25{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)}}{{\left( {970{\rm{ kg}} + 0.25{\rm{ kg}}} \right)}}\\\\ = - 0.52{\rm{ m/}}{{\rm{s}}^2}\\\end{array}

Draw the free body diagram of the cheese block.

W = m2g

Apply Newton’s second law in the vertical direction.

FNm2g=m2a{F_N} - {m_2}g = {m_2}a'

Solve for FN{F_N} .

FN=m2a+m2g=m2(a+g)\begin{array}{c}\\{F_N} = {m_2}a' + {m_2}g\\\\ = {m_2}\left( {a + g} \right)\\\end{array}

Substitute 0.25kg0.25{\rm{ kg}} for m2{m_2} , 0.52m/s2 - 0.52{\rm{ m/}}{{\rm{s}}^2} for aa' and 9.81m/s29.81{\rm{ m/}}{{\rm{s}}^2} for gg .

FN=0.25kg(0.52m/s2+9.81m/s2)=2.32N\begin{array}{c}\\{F_N} = 0.25{\rm{ kg}}\left( { - 0.52{\rm{ m/}}{{\rm{s}}^2} + 9.81{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 2.32{\rm{ N}}\\\end{array}

Ans: Part a

The amount of the work done by the cable on the elevator cab to displace a distance of d1{d_1} is 27.76kJ{\rm{27}}{\rm{.76 kJ}} .

Part b

The magnitude of FN{F_N} is 2.32N2.32{\rm{ N}} .

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