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:
- The tension \( T \) in the upward direction.
- 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:
\[ \boxed{74.5 , \text{N}} \]
2 answers