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Chapter 6: Problem 46

constant eastward acceleration of \(a=2.80 \mathrm{~m} / \mathrm{s}^{2}\). A worker assists the cart by pushing on the crate with a force that is eastward and has magnitude that depends on time according to \(F(t)=(5.40 \mathrm{~N} / \mathrm{s}) t .\) What is the instantaneous power supplied by this force at \(t=5.00 \mathrm{~s} ?\)

Short Answer

Expert verified
The instantaneous power supplied by the force at \( t=5.00s \) is 378.00W.

Step by step solution

01

Calculate Force at the given time

Use the formula for force \( F(t) = 5.40N/s \cdot t \) to find the force at \( t = 5.00s \). This comes out to be \( F(5) = 5.40N/s \cdot 5.00s = 27.00N \).
02

Compute Velocity at the given time

To calculate the velocity, use the equation for velocity at constant acceleration \( v = u + at \). For this case, the initial velocity (u) is 0 (since it's not specified, we assume the object starts from rest). Hence, the velocity at \( t = 5.00s \) is \( v = 0 + 2.80m/s^2 \cdot 5.00s = 14.00m/s \).
03

Compute Instantaneous Power

The instantaneous power is the product of force and velocity which we have already computed in the previous steps. Therefore, substituting these values into the power formula \( P = F \cdot v \) gives us \( P = 27.00N \cdot 14.00m/s = 378.00W \).

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Key Concepts

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

Constant Acceleration
Understanding constant acceleration is crucial for solving problems in mechanics, part of the kinematics branch of physics.

Constant acceleration occurs when an object speeds up or slows down at a steady rate over a period of time. The concept can be encapsulated in the formula: \[ v = u + at \], where:
  • \( v \) is the final velocity,
  • \( u \) is the initial velocity,
  • \( a \) is the constant acceleration, and
  • \( t \) is the time.
When \( u \) is zero, which commonly means the object is initially at rest, the formula simplifies to \( v = at \). In the context of the exercise, the acceleration was constant and eastward, allowing us to calculate the velocity at any given time effortlessly if we know the acceleration and the time elapsed.
Force-Time Dependence
The concept of force-time dependence expresses how a force applied to an object might change over time.

In many physical situations, the force is not constant but varies, often as a function of time. Such a relationship can be depicted as \( F(t) \), highlighting how the magnitude of the force depends on time \( t \). In our exercise, the force applied to the cart has been defined by the equation \( F(t) = (5.40 \, N/s) \cdot t \), indicating that the force increases linearly with time. At any given moment, the force exerted can be calculated by simply inserting the elapsed time into this equation. Applying the equation at \( t = 5.00 \, s \) shows us that the force was \( 27.00 \, N \) at that specific instant, highlighting the dynamic nature of force-time dependent phenomena.
Kinematics
The branch of physics known as kinematics delves into the motion of objects without considering the forces which cause such movement.

It deals with concepts like velocity, acceleration, displacement, and time. These relationships are crucial for understanding motion and are foundational for more complex physics problems. The kinematic equations enable us to link these quantities and solve for unknowns, given certain initial conditions. An elementary equation for velocity is the same we encountered in the context of constant acceleration: \[ v = u + at \]. The notions of kinematics are applicable in our exercise, allowing for the calculation of an object's instantaneous velocity at a given time, which can later be used to find the instantaneous power. Kinematics provides a straightforward approach to analyzing motion that is independent of the forces that cause it, which is key to dissecting many real-world problems.

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Most popular questions from this chapter

Using a cable with a tension of \(1350 \mathrm{~N}\), a tow truck pulls a car \(5.00 \mathrm{~km}\) along a horizontal roadway. (a) How much work does the cable do on the car if it pulls horizontally? If it pulls at \(35.0^{\circ}\) above the horizontal? (b) How much work does the cable do on the tow truck in both cases of part (a)? (c) How much work does gravity do on the car in part (a)?

The upper end of a light rope of length \(L=0.600 \mathrm{~m}\) is attached to the ceiling, and a small steel ball with mass \(m=0.200 \mathrm{~kg}\) is suspended from the lower end of the rope. Initially the ball is at rest and the rope is vertical. Then a force \(\vec{F}\) with constant magnitude \(F=0.760 \mathrm{~N}\) and a direction that is maintained tangential to the path of the ball is applied and the ball moves in an arc of a circle of radius \(L\). What is the speed of the ball when the rope makes an angle \(\alpha=37.0^{\circ}\) with the vertical?

To stretch an ideal spring \(3.00 \mathrm{~cm}\) from its unstretched length, \(12.0 \mathrm{~J}\) of work must be done. (a) What is the force constant of this spring? (b) What magnitude force is needed to stretch the spring 3.00 \(\mathrm{cm}\) from its unstretched length? (c) How much work must be done to compress this spring \(4.00 \mathrm{~cm}\) from its unstretched length, and what force is needed to compress it this distance?

Leg Presses. As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff ideal springs arranged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do \(80.0 \mathrm{~J}\) of work when you compress the springs \(0.200 \mathrm{~m}\) from their uncompressed length. (a) What magnitude of force must you apply to hold the platform in this position? (b) How much additional work must you do to move the platform \(0.200 \mathrm{~m}\) farther, and what maximum force must you apply?

A force in the \(+x\) -direction with magnitude \(F(x)=18.0 \mathrm{~N}-(0.530 \mathrm{~N} / \mathrm{m}) x\) is applied to a \(6.00 \mathrm{~kg}\) box that is sitting on the horizontal, frictionless surface of a frozen lake. \(F(x)\) is the only horizontal force on the box. If the box is initially at rest at \(x=0\) what is its speed after it has traveled \(14.0 \mathrm{~m} ?\)
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