Problem 77 - University Physics with Modern Physics

Question

CALC A block of ice with mass 4.00 \(\mathrm{kg}\) is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal force \(\vec{\boldsymbol{F}}\) to it. As a result, the block moves along the \(x\) -axis such that its position as a function of time is given by 3 . \(x(t)=\alpha t^{2}+\beta t^{3},\) where \(\alpha=0.200 \mathrm{m} / \mathrm{s}^{2}\) and \(\beta=0.0200 \mathrm{m} / \mathrm{s}^{3}\). (a) Calculate the velocity of the object when \(t=4.00\) s. (b) Calculate the magnitude of \(F\) when \(t=4.00\) s. (c) Calculate the work done by the force \(\vec{F}\) during the first 4.00 s of the motion.

Step-by-Step Solution

Answer

(a) 2.56 m/s, (b) 3.52 N, (c) 13.1072 J.

Step 1: Position Function and Derivative Setup

First, we need to find the velocity of the block as a function of time. The position of the block as a function of time is given by \( x(t) = \alpha t^2 + \beta t^3 \), where \( \alpha = 0.200 \text{ m/s}^2 \) and \( \beta = 0.0200 \text{ m/s}^3 \). To find the velocity, we take the derivative of the position function with respect to time.

Step 2: Find the Velocity Function

To find the velocity function \( v(t) \), take the derivative of \( x(t) = 0.200 t^2 + 0.0200 t^3 \) with respect to \( t \):\[ v(t) = \frac{d}{dt}(\alpha t^2 + \beta t^3) = 2\alpha t + 3\beta t^2 \]Substitute \( \alpha = 0.200 \) and \( \beta = 0.0200 \):\[ v(t) = 0.400t + 0.060t^2 \]

Step 3: Calculate Velocity at t = 4 s

Substitute \( t = 4.00 \) s into the velocity function \( v(t) = 0.400t + 0.060t^2 \):\[ v(4.00) = 0.400 \times 4.00 + 0.060 \times (4.00)^2 \]\[ v(4.00) = 1.600 + 0.960 = 2.560 \text{ m/s} \] Thus, the velocity of the object at \( t = 4.00 \text{ s} \) is \( 2.560 \text{ m/s} \).

Step 4: Find Acceleration Function

The force applied is related to the acceleration of the block by Newton's Second Law. To find the acceleration, take the derivative of the velocity function \( v(t) = 0.400t + 0.060t^2 \) with respect to \( t \):\[ a(t) = \frac{d}{dt}(0.400t + 0.060t^2) = 0.400 + 0.120t \]

Step 5: Calculate Acceleration at t = 4 s

Substitute \( t = 4.00 \) s into the acceleration function \( a(t) = 0.400 + 0.120t \):\[ a(4.00) = 0.400 + 0.120 \times 4.00 \]\[ a(4.00) = 0.400 + 0.480 = 0.880 \text{ m/s}^2 \] The acceleration at \( t = 4.00 \text{ s} \) is \( 0.880 \text{ m/s}^2 \).

Step 6: Apply Newton's Second Law to Find Force

Using Newton's Second Law, \( F = ma \), where \( m = 4.00 \text{ kg} \) and \( a = 0.880 \text{ m/s}^2 \):\[ F = 4.00 \times 0.880 \]\[ F = 3.52 \text{ N} \] The magnitude of the force \( F \) at \( t = 4.00 \text{ s} \) is \( 3.52 \text{ N} \).

Step 7: Calculate Work Done by Force

The work done by the force over time \( t = 0 \) to \( t = 4.00 \) s can be calculated using the work-energy principle, \( W = \int F dx \). Here, work done (\( W \)) is the change in kinetic energy. First, we find the final kinetic energy:\[ KE_f = \frac{1}{2} m v^2(4.00) \]Substitute \( m = 4.00 \text{ kg} \) and \( v(4.00) = 2.56 \text{ m/s} \): \[ KE_f = \frac{1}{2} \times 4.00 \times (2.56)^2 = 13.1072 \text{ J} \]The initial kinetic energy \( (KE_i) \) is 0 because it starts at rest, \( W = KE_f - KE_i = 13.1072 \text{ J} - 0 = 13.1072 \text{ J} \).

Step 8: Conclusion

Therefore, the velocity of the object at \( t = 4.00 \) s is \( 2.56 \text{ m/s} \), the force magnitude at \( t = 4.00 \) s is \( 3.52 \text{ N} \), and the work done by the force during the first 4.00 seconds is \( 13.1072 \text{ J} \).

Key Concepts

Kinematics

Kinematics studies how objects move. It focuses on the description of motion, primarily using equations that relate position, velocity, and acceleration. In our exercise, the position of the block as a function of time is given by\[ x(t) = \alpha t^2 + \beta t^3 \]where the constants \(\alpha\) and \(\beta\) represent specific values tied to the object's motion. To find the velocity, we differentiate this position function with respect to time. When you differentiate, you are essentially calculating how fast or slow a quantity is changing. This derivative provides a velocity function \(v(t)\):- Velocity tells us how fast the object is moving and in which direction.
- It gives the rate of change of position with respect to time.Once we have the velocity function, we can determine the acceleration by differentiating velocity. Acceleration is another critical kinematic concept:- It describes the rate at which velocity changes with time.
- This can involve picking up speed (positive acceleration) or slowing down (negative acceleration).
This step-by-step process helps solve for not just how objects move but the characteristics of their motion.

Newton's Laws

Newton's Laws of Motion form the cornerstone of classical mechanics, describing the relationship between a body and the forces acting upon it. The second law is essential here, as it links force, mass, and acceleration:\[ F = ma \]This equation states that force is the product of mass and acceleration. In the problem, the worker applies a horizontal force to the ice block which then starts moving. Here's how this law applies:- **Mass** (\(m\)) is a measure of how much matter the block contains (here, it's 4.00 kg).
- **Acceleration** (\(a\)) can be determined from the velocity function already established.
- **Force** (\(F\)) is the interaction that changes the motion of the block.
The steps in the solution walk through calculating the force magnitude by substituting the known mass and calculated acceleration at a specific time (\(t = 4 \space \text{s}\)). Newon's laws help explain why and how the block accelerates when a net external force acts on it, exemplifying cause and effect in motion.

Work and Energy

Work and energy principles give insights into how forces doing work affect an object's motion and energy. In this context, understanding how work is done helps us calculate changes in kinetic energy.The work-energy principle states:- **Work** (\(W\)) is the product of the force applied to an object and the distance over which it acts. In this case: - The work done can be linked to the change in kinetic energy.Kinetic energy \((KE)\) represents the energy of motion:- **Initial Kinetic Energy** (\(KE_i\)) of the ice block is zero since it starts from rest.
- **Final Kinetic Energy** (\(KE_f\)) is calculated at a later time using: \[ KE = \frac{1}{2} m v^2 \]By understanding work done on the block, we ascertain how the force's application over the first 4 seconds changes the block's state of motion. Thus, work done not only gives us a quantitative measure of energy transformation but highlights how forces bring about motions and alterations in speed/position.