Problem 73 - Physics for Scientists and Engineers with Modern Physics

Question

(III) The position of a 280 -g object is given (in meters) by \(x=5.0 t^{3}-8.0 t^{2}-44 t,\) where \(t\) is in seconds. Determine the net rate of work done on this object \((a)\) at \(t=2.0 \mathrm{~s}\) and (b) at \(t=4.0 \mathrm{~s}\). (c) What is the average net power input during the interval from \(t=0 \mathrm{~s}\) to \(t=2.0 \mathrm{~s},\) and in the interval from \(t=2.0 \mathrm{~s}\) to \(4.0 \mathrm{~s} ?\)

Step-by-Step Solution

Answer

(a) Net power at \( t=2s \): \( P(2.0) \); (b) at \( t=4s \): \( P(4.0) \); (c) Avg. net power: Interval [0,2] and [2,4].

Step 1: Differentiate Position to Find Velocity

The position function is given by \( x(t) = 5.0t^3 - 8.0t^2 - 44t \). To find velocity, differentiate this with respect to time, \( v(t) = \frac{dx}{dt} = 15.0t^2 - 16.0t - 44 \).

Step 2: Differentiate Velocity to Find Acceleration

Now differentiate the velocity function \( v(t) = 15.0t^2 - 16.0t - 44 \) to get acceleration \( a(t) = \frac{dv}{dt} = 30.0t - 16.0 \).

Step 3: Calculate Force Using Acceleration

Since the mass \( m = 280 \) g \( = 0.28 \) kg, the force is \( F(t) = m \cdot a(t) = 0.28(30.0t - 16.0) = 8.4t - 4.48 \) N.

Step 4: Calculate Power as Function of Time

Power is given by the product of force and velocity: \( P(t) = F(t) \cdot v(t) = (8.4t - 4.48)(15.0t^2 - 16.0t - 44) \). Simplify to find \( P(t) \).

Step 5: Evaluate Power at Specific Times

(a) Substitute \( t = 2.0 \) s and (b) \( t = 4.0 \) s into \( P(t) \) to get power at these times. Solve for \( P(2.0) \) and \( P(4.0) \).

Step 6: Calculate Work Done over Given Intervals

Work done over an interval \([t_1, t_2]\) is the integral of power: \( W = \int_{t_1}^{t_2} P(t) \, dt \). Compute \( W_{0 \to 2} \) and \( W_{2 \to 4} \).

Step 7: Find Average Net Power Input

Average power input is work done divided by time interval: \( \text{Average Power} = \frac{W}{\text{time interval}} \). Calculate for intervals \( [0, 2] \) and \( [2, 4] \).

Key Concepts

Kinematics

In physics, understanding motion starts with kinematics, which is all about describing how objects move. In this exercise, we deal with an object's movement on a straight line, described by a mathematical function. The position function, given as \( x(t) = 5.0t^{3} - 8.0t^{2} - 44t \), helps us track where the object is over time. By finding its derivative, we grasp how fast the object moves, which is called velocity.
Velocity involves having both direction and speed, making it a vector quantity. By differentiating the position function with respect to time, we obtain the velocity function \( v(t) = 15.0t^{2} - 16.0t - 44 \). This derivative tells us the object’s speed and direction at any given time. Understanding velocity is crucial because it forms the bridge to further concepts like acceleration, which we'll look into next.
Kinematics doesn’t stop at positions and velocities. The next step is finding velocity's rate of change, namely acceleration, which offers deeper insight into the motion. Acceleration is found by differentiating the velocity function \( v(t) \) itself, giving us \( a(t) = 30.0t - 16.0 \). This process shows how kinematics serves as the foundation for understanding motion through calculations involving position, velocity, and acceleration.

Dynamics

Dynamics comes into play when we introduce forces affecting motion. Once we know the acceleration, we leverage Newton's second law to determine the force acting on the object. Newton’s second law tells us that force is the product of mass and acceleration: \( F = m imes a \). In our problem, the mass \( m \) is 0.28 kg, as the object weighs 280 grams.
By using the formula \( F(t) = m \cdot a(t) = 0.28(30.0t - 16.0) \), we get the force function \( F(t) = 8.4t - 4.48 \) N. This expression allows us to determine the force acting on the object over time based on its acceleration. Understanding dynamics is critical because it tells us not just what is happening—like in kinematics—but why. It explains the underlying causes that bring about motion, which are forces.
Moreover, dynamics is interconnected with other mechanical phenomena, like work and power, discussed in the next section. Remember, without dynamics, our understanding of motion would stay incomplete as it doesn't consider the influence of external forces.

Work and Power

Work and power give insight into how energy is transferred and utilized when a force acts over a distance. In this exercise, work done by a force, and how fast this work is done (power), is calculated. Whenever a force causes a displacement, work is done according to the formula \( W = F \times d \) when force is constant. But when dealing with variable forces, like in our exercise, we use integrals to calculate work over a time interval.
Power, on the other hand, measures how quickly work is done. In our context, power \( P(t) \) is the product of force and velocity, \( P(t) = F(t) \times v(t) \), which mathematically is \( P(t) = (8.4t - 4.48)(15.0t^2 - 16.0t - 44) \). We substitute specific times into this equation to find the power at \( t = 2 \) s and \( t = 4 \) s, releasing insights into how the rate of energy transfer evolves with time.
Finally, understanding what average power implies is crucial. It is the total work done over a period divided by the time this took. This provides an average measure of how efficiently energy is exerted over a time span. Calculating it over different intervals, such as from 0 to 2 seconds and from 2 to 4 seconds, allows us to compare energy usage rates across different phases, which is crucial for examining mechanical systems.