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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

A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from \(x=0\) to \(x=6.9 \mathrm{~m}\) as you apply a force with \(x\) -component \(F_{x}=-[20.0 \mathrm{~N}+(3.0 \mathrm{~N} / \mathrm{m}) x] .\) How much work does the force you apply do on the cow during this displacement?

One end of a horizontal spring with force constant \(76.0 \mathrm{~N} / \mathrm{m}\) is attached to a vertical post. A \(2.00 \mathrm{~kg}\) block of frictionless ice is attached to the other end and rests on the floor. The spring is initially neither stretched nor compressed. A constant horizontal force of 54.0 \(\mathrm{N}\) is then applied to the block, in the direction away from the post. (a) What is the speed of the block when the spring is stretched \(0.400 \mathrm{~m} ?\) (b) At that instant, what are the magnitude and direction of the acceleration of the block?

A 12.0 kg package in a mail-sorting room slides \(2.00 \mathrm{~m}\) down a chute that is inclined at \(53.0^{\circ}\) below the horizontal. The coefficient of kinetic friction between the package and the chute's surface is 0.40 . Calculate the work done on the package by (a) friction, (b) gravity, and (c) the normal force. (d) What is the net work done on the package?

A luggage handler pulls a \(20.0 \mathrm{~kg}\) suitcase up a ramp inclined at \(32.0^{\circ}\) above the horizontal by a force \(\vec{F}\) of magnitude \(160 \mathrm{~N}\) that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is \(\mu_{\mathrm{k}}=0.300 .\) If the suitcase travels \(3.80 \mathrm{~m}\) along the ramp, calculate (a) the work done on the suitcase by \(\overrightarrow{\boldsymbol{F}} ;\) (b) the work done on the suitcase by the gravitational force; (c) the work done on the suitcase by the normal force; (d) the work done on the suitcase by the friction force; (e) the total work done on the suitcase. (f) If the speed of the suitcase is zero at the bottom of the ramp, what is its speed after it has traveled \(3.80 \mathrm{~m}\) along the ramp?

A physics professor is pushed up a ramp inclined upward at \(30.0^{\circ}\) above the horizontal as she sits in her desk chair, which slides on frictionless rollers. The combined mass of the professor and chair is \(85.0 \mathrm{~kg} .\) She is pushed \(2.50 \mathrm{~m}\) along the incline by a group of students who together exert a constant horizontal force of \(600 \mathrm{~N}\). The professor's speed at the bottom of the ramp is \(2.00 \mathrm{~m} / \mathrm{s}\). Use the work-energy theorem to find her speed at the top of the ramp.
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