Question

The figure shows the velocity graph of a 75 kg passenger in an elevator.

1)What is the passenger's apparent weight at  t=1.0s?

2)What is the passenger's apparent weight at t = 5.0s?

3)What is the passenger's apparent weight at t = 9.0s?

Answer 1

Concepts and reason

The concepts required to solve the problem are acceleration, and Newton’s second law.

From the slope of graph, calculate the acceleration of body. Then calculate the apparent weight of passenger at different time instants using the Newton’s second law.

Fundamentals

The acceleration of an object is defined as the rate of change of velocity. It is also a vector quantity.

It is also calculated from the velocity-time graph. The slope of v-t graph gives us the acceleration of object.

The force from the Newton’s second law force is defined as the product of mass of an object and the acceleration of an object.

The weight of an object is defined as the gravitational force exerted on the object of mass m.

The expression to find out the acceleration of an object from v-t graph is given as follows:

$\begin{array}{c}\\a = \frac{{{\rm{change in velocity}}}}{{{\rm{change in time}}}}\\\\ = \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}\\\\ = \frac{{\Delta v}}{{\Delta t}}\\\end{array}$

The expression of force from Newton’s second law is given as follows:

$F = ma$

The weight of an object is given as follows:

$W = mg$

(1)

The acceleration of passenger is given as follows:

$a = \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}$

Here, ${v_2}$is the velocity at time ${t_2}$ and ${v_1}$ is the velocity at time ${t_1}$.

Substitute 8.0 m/sec for ${v_2}$, 0.0 m/sec for ${v_1}$, 2.0 sec for ${t_2}$ and 0.0 sec for ${t_1}$ in the expression $a = \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}$.

$\begin{array}{c}\\a = \frac{{{\rm{8}}{\rm{.0 m/sec}} - 0.0{\rm{ m/sec}}}}{{{\rm{2}}{\rm{.0 sec}} - 0.0{\rm{ sec}}}}\\\\ = 4{\rm{ m/se}}{{\rm{c}}^2}\\\end{array}$

Consider that the elevator is going up and acceleration is positive.

The expression of the passenger’s apparent weight is calculated as follows:

$\begin{array}{c}\\\sum {F = ma} \\\\N - mg = ma\\\\N = m\left( {a + g} \right)\\\end{array}$

Here, N is the normal reaction, m is the mass, a is the acceleration and g is the acceleration due to gravity.

Substitute 75 kg for m, $4{\rm{ m/se}}{{\rm{c}}^2}$ for a and $9.8{\rm{ m/se}}{{\rm{c}}^2}$ for g in the expression $N = m\left( {a + g} \right)$.

$\begin{array}{c}\\N = 75{\rm{ kg}}\left( {9.8{\rm{ m/se}}{{\rm{c}}^2} + 4{\rm{ m/se}}{{\rm{c}}^2}} \right)\\\\ = 1035{\rm{ N}}\\\end{array}$

(2)

Substitute 8.0 m/sec for ${v_2}$, 8.0 m/sec for ${v_1}$, 6.0 sec for ${t_2}$ and 2.0 sec for ${t_1}$ in the expression $a = \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}$.

$\begin{array}{c}\\a = \frac{{{\rm{8}}{\rm{.0 m/sec}} - 8.0{\rm{ m/sec}}}}{{{\rm{6}}{\rm{.0 sec}} - 2.0{\rm{ sec}}}}\\\\ = 0{\rm{ m/se}}{{\rm{c}}^2}\\\end{array}$

In this case velocity is not changing implies that the acceleration is zero.

The expression of the passenger’s apparent weight is calculated as follows:

$\begin{array}{c}\\\sum {F = 0} \\\\N - mg = 0\\\\N = mg\\\end{array}$

Here, N is the normal reaction, m is the mass and g is the acceleration due to gravity.

Substitute 75 kg for m and $9.8{\rm{ m/se}}{{\rm{c}}^2}$ for g in the expression $N = mg$.

$\begin{array}{c}\\N = 75{\rm{ kg}}\left( {9.8{\rm{ m/se}}{{\rm{c}}^2}} \right)\\\\ = 735{\rm{ N}}\\\end{array}$

(3)

The acceleration of passenger is given as follows:

$a = \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}$

Here, ${v_2}$is the velocity at time ${t_2}$ and ${v_1}$ is the velocity at time ${t_1}$.

Substitute 0.0 m/sec for ${v_2}$, 8.0 m/sec for ${v_1}$, 6.0 sec for ${t_2}$ and 6.0 sec for ${t_1}$ in the expression $a = \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}$.

$\begin{array}{c}\\a = \frac{{{\rm{0}}{\rm{.0 m/sec}} - 8.0{\rm{ m/sec}}}}{{{\rm{10}}{\rm{.0 sec}} - 6.0{\rm{ sec}}}}\\\\ = - 2{\rm{ m/se}}{{\rm{c}}^2}\\\end{array}$

Consider that the elevator is going up and acceleration is negative.

The expression of the passenger’s apparent weight is calculated as follows:

$\begin{array}{c}\\\sum {F = - ma} \\\\N - mg = - ma\\\\N = m\left( {g - a} \right)\\\end{array}$

Here, N is the normal reaction, m is the mass, a is the acceleration and g is the acceleration due to gravity.

Substitute 75 kg for m, ${\rm{2 m/se}}{{\rm{c}}^2}$ for a and $9.8{\rm{ m/se}}{{\rm{c}}^2}$ for g in the expression $N = m\left( {g - a} \right)$.

$\begin{array}{c}\\N = 75{\rm{ kg}}\left( {9.8{\rm{ m/se}}{{\rm{c}}^2} - 2{\rm{ m/se}}{{\rm{c}}^2}} \right)\\\\ = 585{\rm{ N}}\\\end{array}$

Ans: Part 1

The passenger’s apparent weight is 1035 N.

Part 2

The passenger’s apparent weight is 735 N.

Part 3

The passenger’s apparent weight is 585 N.