Question

A crate on a motorized cart starts from rest and moves with a constant eastward acceleration of \(a = 2.80 \, \mathrm{m/s}^2\). 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 \(F(t) = (5.40 \, \mathrm{N/s})t\). What is the instantaneous power supplied by this force at \(t = 5.00\) s?

Short answer

The power at \(t = 5.00\) s is 378 W.

Step-by-step solution

Identify Variables and Equations

Given the acceleration \(a = 2.80 \, \mathrm{m/s}^2\) and force as a function of time \(F(t) = (5.40 \, \mathrm{N/s})t\). We want to find the instantaneous power \(P\) at \(t = 5.00\) s. The formula for power is \(P = F \cdot v\), where \(F\) is force and \(v\) is the velocity of the crate at that instant.

Determine Force at t = 5.00 s

Substitute \(t = 5.00\) s into the force equation \(F(t) = (5.40 \, \mathrm{N/s})t\). This gives us \(F(5.00) = (5.40 \, \mathrm{N/s})(5.00 \, \mathrm{s}) = 27.00 \, \mathrm{N}\).

Calculate Velocity at t = 5.00 s

The cart starts from rest, so initial velocity \(v_0 = 0\). Use the formula \(v = v_0 + at\) to find the velocity. With \(a = 2.80 \, \mathrm{m/s}^2\) and \(t = 5.00 \, \mathrm{s}\), we have \(v = 0 + (2.80 \, \mathrm{m/s}^2)(5.00 \, \mathrm{s}) = 14.00 \, \mathrm{m/s}\).

Compute Instantaneous Power

Use the power formula \(P = F \cdot v\) and plug in the values \(F = 27.00 \, \mathrm{N}\) and \(v = 14.00 \, \mathrm{m/s}\):\[P = 27.00 \, \mathrm{N} \times 14.00 \, \mathrm{m/s} = 378.00 \, \mathrm{W}\].

Key concepts

Constant Acceleration

Constant acceleration refers to a situation where the acceleration of an object does not change over time. In this problem, the crate is moving with a constant eastward acceleration of \(a = 2.80 \, \mathrm{m/s}^2\). This means that the crate’s velocity increases by 2.80 meters per second every second.

Knowing the acceleration helps us predict how fast the crate is moving at any given moment. For instance, if the crate started from rest, its velocity after a time \(t\) can be calculated with the simple formula:

• \(v = v_0 + at\)

where \(v_0\) is the initial velocity. Here, \(v_0 = 0\), because the crate starts from rest. Understanding constant acceleration is crucial in physics to foresee the future motion of objects.

Force as a Function of Time

In this exercise, the force applied by a worker on the crate isn't constant. Instead, it changes with time. This is expressed by the function \(F(t) = (5.40 \, \mathrm{N/s})t\). It's a linear relation telling us that the force increases at a rate of \(5.40 \, \mathrm{Newtons}\) per second.

At any specific time \(t\), the magnitude of the force can be calculated by multiplying 5.40 by that time \(t\).

• For example, at \(t = 5.00 \, \mathrm{s}\), substituting into the function gives force \(F(5.00) = 27.00 \, \mathrm{N}\).

Understanding how force varies with time is essential in determining how it influences the motion of objects in real-time scenarios.

Velocity Calculation

Velocity represents how fast something is moving in a specific direction. For the crate, the velocity at any given moment depends on two factors: how long it has been accelerating and the rate of acceleration.

Starting from rest with a constant acceleration, the velocity can be calculated using the formula:

• \(v = v_0 + at\)

Here, \(v_0\) (initial velocity) is \(0\), as indicated by the problem statement. For \(t = 5.00 \, \mathrm{s}\), the calculation becomes \(v = 0 + (2.80 \, \mathrm{m/s}^2)(5.00 \, \mathrm{s}) = 14.00 \, \mathrm{m/s}\).

This means after 5 seconds, the crate moves east at 14 meters per second. It shows how constant acceleration impacts velocity over time.

Physics Problem Solving

Solving physics problems effectively requires a methodical approach. To tackle the problem of instantaneous power, we utilize known equations and systematically solve for unknowns.

Steps involved in this exercise include:

• Identifying variables and equations: Recognize given values like acceleration \(a = 2.80 \, \mathrm{m/s}^2\) and force function \(F(t)\).
• Calculate force: Plug the given time into the force function to find the instantaneous force at time \(t\).
• Determine velocity: Use the velocity formula to calculate speed at the specified time.
• Compute power: Finally, multiply force and velocity to get instantaneous power \(P = F \cdot v\).

Approaching such problems requires attention to the flow of information and logical application of physical laws.