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Magazine Chapter 15: Problem 28
The 20 -kg crate is lifted by a force of \(F=\left(100+5 t^{2}\right) \mathrm{N},\) where \(t\) is in seconds. Determine how high the crate has moved upward when \(t=3\) s, starting from rest.
Short Answer
Expert verified
The crate has moved upward by approximately 23.19 meters after 3 seconds.
Step by step solution
01
Determine the acceleration using Newton鈥檚 second law
Newton's second law states that the force acting on an object is equal to its mass times its acceleration (F = ma). We can rearrange that to solve for acceleration (a = F / m). So, the acceleration a(t) of the crate at any time t is given by \[a(t) = \frac{F(t)}{m} = \frac{(100 + 5t^2) N}{20 kg} = 5 + \frac{t^2}{4} m/s^2\].
02
Compute the velocity by integrating the acceleration
To obtain the velocity, we need to integrate the acceleration with respect to time. The indefinite integral of a(t) is \[v(t) = \int a(t) dt = \int (5 + \frac{t^2}{4}) dt = 5t + \frac{t^3}{12} + C1\]. Because the crate starts from rest, the initial velocity is zero and thus C1 = 0, which simplifies the equation to \[v(t) = 5t + \frac{t^3}{12}\].
03
Compute the displacement by integrating the velocity
We now need to find the position as a function of time. This involves taking the integral of the velocity function, which gives \[h(t) = \int v(t) dt = \int (5t + \frac{t^3}{12}) dt = \frac{5t^2}{2} + \frac{t^4}{48} + C2\]. Considering that the initial position is zero (the crate starts on the ground), this means that C2 = 0 and our equation becomes \[h(t) = \frac{5t^2}{2} + \frac{t^4}{48}\].
04
Calculate the displacement at \(t = 3\)s
Finally, we substitute \(t = 3\)s into the above expression to find the crate's upward displacement at that time, \[h(3) = \frac{5(3)^2}{2} + \frac{(3)^4}{48} = 23.1875 m\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Acceleration
Acceleration is a core concept in understanding motion. It refers to how an object's velocity changes over time. In this exercise, we have a 20-kg crate being lifted by a force that changes with time:
Here, the acceleration is not constant because it relies on time due to the time-dependent force. This equation shows how fast the crate speeds up at any moment as it goes up, and understanding this helps us explore how the velocity and displacement change too.
- The force is given as a function of time: \(F(t) = 100 + 5t^2\).
- According to Newton's Second Law, force is equal to the mass of an object times its acceleration (\(F = ma\)).
Here, the acceleration is not constant because it relies on time due to the time-dependent force. This equation shows how fast the crate speeds up at any moment as it goes up, and understanding this helps us explore how the velocity and displacement change too.
Velocity Integration
To find the velocity from acceleration, we use a process called integration. Velocity tells us how quickly the position of an object changes over time.
This equation means that as time progresses, the velocity of the crate increases not only because of the constant acceleration but also because t-squared contribution increases, making it speed up faster over time.
- The expression for acceleration, \(a(t) = 5 + \frac{t^2}{4}\), needs to be integrated with respect to time \(t\).
- The integration gives us the velocity function: \[v(t) = \int (5 + \frac{t^2}{4}) dt = 5t + \frac{t^3}{12} + C1\].
This equation means that as time progresses, the velocity of the crate increases not only because of the constant acceleration but also because t-squared contribution increases, making it speed up faster over time.
Displacement Calculation
Determining how far the crate travels involves integrating the velocity equation over time. Displacement tells us the change in position of the crate.
By substituting \(t = 3\) seconds into this equation, we can evaluate how high the crate travels, which turns out to be approximately \(23.1875\) meters. This calculation shows that knowing velocity helps us find how far the object actually moves.
- Our velocity function is \(v(t) = 5t + \frac{t^3}{12}\).
- Integrating this with respect to time gives the displacement equation: \[h(t) = \int (5t + \frac{t^3}{12}) dt = \frac{5t^2}{2} + \frac{t^4}{48} + C2\].
By substituting \(t = 3\) seconds into this equation, we can evaluate how high the crate travels, which turns out to be approximately \(23.1875\) meters. This calculation shows that knowing velocity helps us find how far the object actually moves.
Force Analysis
Analyzing the forces acting on an object can help us understand the motion thoroughly. In this exercise:
This force analysis is crucial because it gives insight into how varying forces influence an object's motion and contributes to changes in velocity and displacement over time.
- The lifting force is expressed as a function of time: \(F(t) = 100 + 5t^2\).
- This represents an external force applied to the crate causing it to accelerate upwards.
- The crate's weight, a force due to gravity (\(mg = 20 \times 9.8 = 196 \text{ N}\)), also acts on it, but in the opposite direction to the lifting force.
This force analysis is crucial because it gives insight into how varying forces influence an object's motion and contributes to changes in velocity and displacement over time.
