An elevator starts from rest with a constant

upward acceleration and moves 1 m in the first
1.7 s. A passenger in the elevator is holding a
7.1 kg bundle at the end of a vertical cord.
What is the tension in the cord as the elevator accelerates? The acceleration of gravity
is 9.8 m/s2
.
Answer in units of

1 answer

To find the tension in the cord while the elevator is accelerating, we first need to determine the acceleration of the elevator.

Given that the elevator starts from rest and moves 1 meter in 1.7 seconds, we can use the following kinematic equation that describes motion with constant acceleration:

\[ d = v_0 t + \frac{1}{2} a t^2 \]

where:

  • \( d \) is the distance moved (1 m),
  • \( v_0 \) is the initial velocity (0 m/s, since it starts from rest),
  • \( a \) is the acceleration,
  • \( t \) is the time (1.7 s).

Substituting the values we have:

\[ 1 m = 0 + \frac{1}{2} a (1.7 s)^2 \]

This simplifies to:

\[ 1 = \frac{1}{2} a (2.89) \] \[ 1 = 1.445 a \] \[ a = \frac{1}{1.445} \approx 0.692 , \text{m/s}^2 \]

Now that we have the acceleration of the elevator (\( a \approx 0.692 , \text{m/s}^2 \)), we can find the tension \( T \) in the cord that supports the 7.1 kg bundle.

The forces acting on the bundle are:

  1. The tension \( T \) in the upward direction.
  2. The weight \( W \) of the bundle acting downward, which is given by \( W = mg \), where \( m = 7.1 , \text{kg} \) and \( g = 9.8 , \text{m/s}^2 \).

Calculating the weight:

\[ W = 7.1 , \text{kg} \times 9.8 , \text{m/s}^2 = 69.58 , \text{N} \]

The net force \( F_{net} \) acting on the bundle can be expressed as:

\[ F_{net} = T - W \]

According to Newton's second law, \( F_{net} = ma \):

\[ ma = T - W \]

Substituting in the known values:

\[ (7.1 , \text{kg})(0.692 , \text{m/s}^2) = T - 69.58 , \text{N} \]

Calculating the left side:

\[ 4.9132 , \text{N} = T - 69.58 , \text{N} \]

Now solve for \( T \):

\[ T = 4.9132 , \text{N} + 69.58 , \text{N} \] \[ T = 74.4932 , \text{N} \]

Rounding to three significant figures, the tension in the cord is approximately:

\[ \boxed{74.5 , \text{N}} \]