Question

a 5.40 kg bucket of water is accelerated upward by a cord of negligible mass whose breaking strength is 85.0 n. if the bucket starts from rest, what is the minimum time required to raise the bucket a vertical distance of 12.0 m without breaking the cord? express your answer with the appropriate units.

Explanation:

Step1: Find the acceleration using Newton's second law

The forces on the bucket are the tension \( T \) upward and weight \( mg \) downward. Newton's second law: \( T - mg = ma \).
Given \( m = 5.40 \, \text{kg} \), \( T = 85.0 \, \text{N} \), \( g = 9.8 \, \text{m/s}^2 \).
Solve for \( a \):
\( a=\frac{T - mg}{m}=\frac{85.0 - 5.40\times9.8}{5.40}=\frac{85.0 - 52.92}{5.40}=\frac{32.08}{5.40}\approx5.94 \, \text{m/s}^2 \).

Step2: Use kinematic equation for distance

The bucket starts from rest (\( u = 0 \)), distance \( s = 12.0 \, \text{m} \), acceleration \( a \approx5.94 \, \text{m/s}^2 \).
Kinematic equation: \( s = ut+\frac{1}{2}at^2 \). Since \( u = 0 \), \( s=\frac{1}{2}at^2 \).
Solve for \( t \):
\( t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2\times12.0}{5.94}}=\sqrt{\frac{24.0}{5.94}}\approx\sqrt{4.04}\approx2.01 \, \text{s} \).

Answer:

\( t_{\text{min}} \approx \boldsymbol{2.01 \, \text{s}} \)