Question

At $t=1.0\mathrm{s}$, a $0.40-\mathrm{kg}$ object is falling with a speed of $6.0\mathrm{m}/\mathrm{s}$. At $t=2.0\mathrm{s}$, it has a kinetic energy of $25\mathrm{J}$. (a) What is the kinetic energy of the object at $t=1.0\mathrm{s}$ ? (b) What is the speed of the object at $t=2.0\mathrm{s}$ ? (c) How much work was done on the object between $t=1.0\mathrm{s}$ and $t=2.0\mathrm{s}$ ?

Short answer

(a) 7.2 J; (b) 11.18 m/s; (c) 17.8 J

Step-by-step solution

Calculate Initial Kinetic Energy (t=1.0 s)

To find the kinetic energy at $t=1.0\mathrm{s}$, use the formula for kinetic energy $KE=\frac{1}{2}m{v}^2$. Here, $m=0.40\mathrm{kg}$ and $v=6.0\mathrm{m}/\mathrm{s}$:$$KE=\frac{1}{2}\times 0.40\times (6.0{)}^2=0.20\times 36=7.2\mathrm{J}$$

Calculate Speed at t=2.0 s from Kinetic Energy

The kinetic energy at $t=2.0\mathrm{s}$ is given as $25\mathrm{J}$. To find the speed, use the formula rearranged for speed:$$KE=\frac{1}{2}m{v}^2\Rightarrow v=\sqrt{\frac{2\times KE}{m}}$$Substituting values $KE=25\mathrm{J}$ and $m=0.40\mathrm{kg}$:$$v=\sqrt{\frac{2\times 25}{0.40}}=\sqrt{\frac{50}{0.40}}=\sqrt{125}=11.18\mathrm{m}/\mathrm{s}$$

Determine Work Done Between t=1.0 s and t=2.0 s

Work done on the object can be calculated using the change in kinetic energy, $W=K{E}_{final}-K{E}_{initial}$. Substituting the initial kinetic energy at $t=1.0\mathrm{s}$ and final kinetic energy at $t=2.0\mathrm{s}$:$$W=25\mathrm{J}-7.2\mathrm{J}=17.8\mathrm{J}$$

Key concepts

Work-Energy Principle

The work-energy principle is a fundamental concept in physics that connects the ideas of work and energy. It states that the work done on an object is equal to the change in its kinetic energy. This principle helps in understanding how energy is transferred or transformed into motion.
This principle can be written mathematically as:

• Work done (W) = Change in Kinetic Energy (ΔKE)
• ΔKE = KE_{final} - KE_{initial}

Where KE represents the kinetic energy of the object. Kinetic energy is given by the formula, $$KE=\frac{1}{2}m{v}^2$$ where 'm' is mass and 'v' is velocity.
The work-energy principle is crucial when analyzing situations like the one described, where an object gains or loses kinetic energy over time due to external work being done on it. Understanding this principle allows us to calculate how much work was involved in changing the speed of an object as seen from the energy change between two points in time.

Kinematics

Kinematics is the branch of physics that studies motion without considering the forces that cause it. It is all about describing how objects move: their speed, velocity, and acceleration, among other factors.
In the given exercise, at times like t=1.0s and t=2.0s, we can use kinematics to determine certain characteristics of the object's motion:

• Velocity: The speed of an object in a particular direction at a specific time.
• Acceleration: The rate at which an object's velocity changes over time.
• Displacement: The change in the object's position over time.

For instance, even without knowing the forces, we can conclude that the object's velocity changed between t=1.0s and t=2.0s, as evidenced by the change in its kinetic energy. Kinematics helps us characterize this change quantitatively, helping us grasp the concept behind the calculations of changing speed within the solution steps. This aspect links closely with how we approach the speed calculation.

Speed Calculation

Calculating speed is one of the fundamental aspects of analyzing motion, as it helps us understand how fast an object is moving at any given moment.
In the exercise, we start by knowing the speed at t=1.0 s (6.0 m/s). Later, using the given kinetic energy at t=2.0 s, we calculate the new speed using the formula:$$v=\sqrt{\frac{2\times KE}{m}}$$This formula comes from rearranging the kinetic energy equation, which links speed directly to how much energy is in motion. Knowing the mass and kinetic energy, we can backtrack to find the speed without direct measurement.
Our solution continued by using this rearranged formula to find that the speed at t=2.0 s is 11.18 m/s. This calculation clarifies how speed changes when work is done, highlighted by the work-energy principle. It shows the importance of speed in determining the dynamic state of an object through its motion.