Chapter 5: Problem 23
A 2.00 -kg box is moving to the right with speed 9.00 \(\mathrm{m} / \mathrm{s}\) on a horizontal, frictionless surface. At \(t=0\) a horizontal force is applied to the box. The force is directed to the left and has magnitude \(F(t)=\left(6.00 \mathrm{N} / \mathrm{s}^{2}\right) t^{2}\) . (a) What distance does the box move from its position at \(t=0\) before its speed is reduced to zero? (b) If the force continues to be applied, what is the speed of the box at \(t=3.00 \mathrm{s} ?\)
Short Answer
Step by step solution
Understand Initial Conditions
Define Net Force Equation
Write Velocity Function
Find Time When Velocity is Zero
Determine Distance Traveled Before Stopping
Recalculate Speed at t=3.00 s
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Newton's Second Law
Kinematics
- Initial Velocity: The box starts with a velocity of 9.00 m/s.
- Acceleration Function: Since the box is subjected to a time-dependent force, its velocity changes over time. The acceleration was calculated as \( a(t) = 3.00t^2 \text{ m/s}^2 \).
- Velocity Function: To find how the velocity evolves, we integrate the acceleration function:\[ v(t) = \int 3.00t^2 \, dt = t^3 + C \]With the initial condition \( v(0) = 9.00 \text{ m/s} \), we find \( C = 9.00 \, \text{m/s} \).Thus, the velocity function is \( v(t) = t^3 + 9.00 \text{ m/s} \).