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Chapter 5: Problem 22

A \(5.00 \mathrm{~kg}\) crate is suspended from the end of a short vertical rope of negligible mass. An upward force \(F(t)\) is applied to the end of the rope, and the height of the crate above its initial position is given by \(y(t)=(2.80 \mathrm{~m} / \mathrm{s}) t+\left(0.610 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3} .\) What is the magnitude of \(F\) when \(t=4.00 \mathrm{~s} ?\)

Short Answer

Expert verified
The magnitude of force \(F\) when \(t=4.00s\) is approximately \(122.2N\).

Step by step solution

01

Derive the velocity equation

The first derivative of the equation for displacement \(y(t)\) with respect to time \(t\) is the velocity of the object. The given equation for displacement is \( y(t) = (2.80 m/s) t + (0.610 m/s^{3}) t^{3} \). Therefore, the velocity at time \(t\) will be \( v(t) = dy(t) / dt = 2.80 m/s + 3 \cdot (0.610 m/s^{3}) t^{2} = 2.80 m/s + 1.83 m/s^{2} t^{2}\).
02

Derive the acceleration equation

The acceleration of the object at time \(t\) can be found by differentiating the velocity equation \(v(t)\) with respect to time \(t\). Therefore, the acceleration at time \(t\) will be \( a(t) = dv(t) / dt = 2 \cdot 1.83 m/s^{2} t = 3.66 m/s^{3} t \).
03

Find the acceleration at \( t=4s \)

Now substitute \( t=4s \) into the acceleration equation to find the acceleration at time \( t=4s \). Therefore, the acceleration at \( t=4.00s \) is \( a(4.00s) = 3.66 m/s^{3} \cdot 4.00s = 14.64 m/s^{2} \).\n
04

Calculate the total force

The total force \( F \) acting on the crate will be the sum of the force due to gravity, which is \( F_{g} = m \cdot g = 5.00kg \cdot 9.8 m/s^{2} = 49 N \), and the force required to generate the acceleration in the crate, which is \( F_{a} = m \cdot a = 5.00kg \cdot 14.64m/s^{2} = 73.2N \). Therefore, the total force \( F \) is \( F = F_{g} + F_{a} = 49N + 73.2N = 122.2N \).
05

Round the answer

In physics, it is typical to round to the nearest decimal, thus the answer should be approximately 122.2N

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematics
Understanding the movement of the crate involves kinematics, the branch of physics that describes motion. The initial height equation for the crate is given by the function:
  • \( y(t) = (2.80 \, \mathrm{m/s}) t + (0.610 \, \mathrm{m/s^3}) t^3 \)
This equation shows how the height changes over time, integrating linear and cubic terms.
In kinematics:
  • A linear term \((2.80 \, \mathrm{m/s}) t\) represents constant velocity movement.
  • A cubic term \((0.610 \, \mathrm{m/s^3}) t^3\) indicates changing acceleration, showing a more complex motion path.
By analyzing this equation, we gain insight into how the crate moves vertically, offering a clear picture of both constant and variable speed components.
Differentiation
Differentiation is crucial for uncovering velocity and acceleration from the position function.
The process begins with differentiating the position function \( y(t) \) to find velocity \( v(t) \):
  • \( v(t) = \frac{dy}{dt} = 2.80 \, \mathrm{m/s} + 3 \cdot (0.610 \, \mathrm{m/s^3}) t^2 \)
  • This simplifies to \( v(t) = 2.80 \, \mathrm{m/s} + 1.83 \, \mathrm{m/s^2} \cdot t^2 \)
Next, we differentiate the velocity function to obtain acceleration \( a(t) \):
  • \( a(t) = \frac{dv}{dt} = 2 \cdot 1.83 \, \mathrm{m/s^2} \cdot t \)
  • Simplified to \( a(t) = 3.66 \, \mathrm{m/s^3} \cdot t \)
This method reveals how acceleration varies with time, specifically when \( t = 4 \) seconds, providing essential information for calculating force later.
Force Calculation
Newton's Second Law of Motion connects force, mass, and acceleration, forming the basis for calculating force here.
For the crate, the total force required is composed of:
  • The gravitational force \( F_g = m \cdot g = 5.00 \, \mathrm{kg} \cdot 9.8 \, \mathrm{m/s^2} = 49 \, \mathrm{N} \)
  • The force due to acceleration \( F_a = m \cdot a = 5.00 \, \mathrm{kg} \cdot 14.64 \, \mathrm{m/s^2} = 73.2 \, \mathrm{N} \)
By summing these forces:
  • \( F = F_g + F_a = 49 \, \mathrm{N} + 73.2 \, \mathrm{N} = 122.2 \, \mathrm{N} \)
This calculation demonstrates how the specific upward force was determined at \( t = 4 \) seconds, applying principles of physics to a real-world scenario, showcasing the interconnectedness of kinematics and dynamics.

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Most popular questions from this chapter

When a crate with mass \(25.0 \mathrm{~kg}\) is placed on a ramp that is inclined at an angle \(\alpha\) below the horizontal, it slides down the ramp with an acceleration of \(4.9 \mathrm{~m} / \mathrm{s}^{2} .\) The ramp is not friction less. To increase the acceleration of the crate, a downward vertical force \(\overrightarrow{\boldsymbol{F}}\) is applied to the top of the crate. What must \(F\) be in order to increase the acceleration of the crate so that it is \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) ? How does the value of \(F\) that you calculate compare to the weight of the crate?

Two blocks, with masses \(4.00 \mathrm{~kg}\) and \(8.00 \mathrm{~kg},\) are connected by a string and slide down a \(30.0^{\circ}\) inclined plane (Fig. \(\mathbf{P 5 . 9 6}\) ). The coefficient of kinetic friction between the \(4.00 \mathrm{~kg}\) block and the plane is \(0.25 ;\) that between the \(8.00 \mathrm{~kg}\) block and the plane is \(0.35 .\) Calculate (a) the acceleration of each block and (b) the tension in the string. (c) What happens if the positions of the blocks are reversed, so that the \(4.00 \mathrm{~kg}\) block is uphill from the \(8.00 \mathrm{~kg}\) block?

A box with mass \(m\) sits at the bottom of a long ramp that is sloped upward at an angle \(\alpha\) above the horizontal. You give the box a quick shove, and after it leaves your hands it is moving up the ramp with an initial speed \(v_{0}\). The box travels a distance \(d\) up the ramp and then slides back down. When it returns to its starting point, the speed of the box is half the speed it started with; it has speed \(v_{0} / 2 .\) What is the coefficient of kinetic friction between the box and the ramp? (Your answer should depend on only \(\alpha\).)

On the ride "Spindletop" at the amusement park Six Flags Over Texas, people stood against the inner wall of a hollow vertical cylinder with radius \(2.5 \mathrm{~m} .\) The cylinder started to rotate, and when it reached a constant rotation rate of \(0.60 \mathrm{rev} / \mathrm{s},\) the floor dropped about \(0.5 \mathrm{~m}\). The people remained pinned against the wall without touching the floor. (a) Draw a force diagram for a person on this ride after the floor has dropped. (b) What minimum coefficient of static friction was required for the person not to slide downward to the new position of the floor? (c) Does your answer in part (b) depend on the person's mass? (Note: When such a ride is over, the cylinder is slowly brought to rest. As it slows down, people slide down the wall to the floor.

Friction in an Elevator. You are riding in an elevator on the way to the 18 th floor of your dormitory. The elevator is accelerating upward with \(a=1.90 \mathrm{~m} / \mathrm{s}^{2} .\) Beside you is the box containing your new computer; the box and its contents have a total mass of \(36.0 \mathrm{~kg} .\) While the elevator is accelerating upward, you push horizontally on the box to slide it at constant speed toward the elevator door. If the coefficient of kinetic friction between the box and the elevator floor is \(\mu_{\mathrm{k}}=0.32,\) what magnitude of force must you apply?
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