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 F_N = 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 F_N?

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?

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 ${d_2}$ and applying Newton’s second law.

Calculate the magnitude of the new ${F_N}$ 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,

${F_{net}} = ma$

Here, ${F_{net}}$ is the net forces acting on the object, $m$ is the mass of the object and $a$ is the acceleration of the object.

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

$W = mg$

Here, $W$ is the weight of the object and $g$ is the acceleration due to gravity.

The expression of the work done is written as,

$W = Fd\cos \theta$

Here, $W$ is the work done, $F$ is the force, $d$ is the displacement and $\theta$ 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.

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

Apply Newton’s second law in the vertical direction.

${F_N} - {m_2}g = {m_2}a$

Rearrange for $a$ .

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

Substitute $3.11{\rm{ N}}$ for ${F_N}$ , $0.25{\rm{ kg}}$ for ${m_2}$ and $9.81{\rm{ m/}}{{\rm{s}}^2}$ for $g$ .

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

Draw the free body diagram of the elevator cab.

In the above diagram, ${m_1}$ is the mass of the elevator, $\left( {{m_1} + {m_2}} \right)$ is the total mass of the elevator, $W$ is the total weight of the elevator and $T$ is the tension in the cable.

Apply Newton’s second law in the vertical direction.

$T - \left( {{m_1} + {m_2}} \right)g = \left( {{m_1} + {m_2}} \right)a$

Rearrange for $T$ .

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

Substitute $970{\rm{ kg}}$ for ${m_1}$ , $0.25{\rm{ kg}}$ for ${m_2}$ , $2.63{\rm{ m/}}{{\rm{s}}^2}$ for $a$ and $9.81{\rm{ m/}}{{\rm{s}}^2}$ for $g$ .

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

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

$\theta = 0^\circ$

Here, $\theta$ 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 ${d_1}$ .

$W{D_1} = T{d_1}\cos \theta$

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

Substitute $12069.91{\rm{ N}}$ for $T$ , $2.3{\rm{ m}}$ for ${d_1}$ and $0^\circ$ for $\theta$ .

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

(b)

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

$W{D_2} = T'{d_2}\cos \theta$

Here, $W{D_2}$ is net work done by the cable to displace the elevator cab at a distance of ${d_2}$ and $T'$ is the new tension in the cable.

Rearrange for $T'$ .

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

Substitute $91.89{\rm{ kJ}}$ for $W{D_2}$ , $10.2{\rm{ m}}$ for ${d_2}$ and $0^\circ$ for $\theta$ .

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

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

Apply Newton’s second law in the vertical direction.

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

Rearrange for $a$ .

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

Substitute $970{\rm{ kg}}$ for ${m_1}$ , $0.25{\rm{ kg}}$ for ${m_2}$ , $9008.82{\rm{ N}}$ for $T'$ and $9.81{\rm{ m/}}{{\rm{s}}^2}$ for $g$ .

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

Draw the free body diagram of the cheese block.

Apply Newton’s second law in the vertical direction.

${F_N} - {m_2}g = {m_2}a'$

Solve for ${F_N}$ .

$\begin{array}{c}\\{F_N} = {m_2}a' + {m_2}g\\\\ = {m_2}\left( {a + g} \right)\\\end{array}$

Substitute $0.25{\rm{ kg}}$ for ${m_2}$ , $- 0.52{\rm{ m/}}{{\rm{s}}^2}$ for $a'$ and $9.81{\rm{ m/}}{{\rm{s}}^2}$ for $g$ .

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

Ans: Part a

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

Part b

The magnitude of ${F_N}$ is $2.32{\rm{ N}}$ .