Question

A man throws a rock of mass \(m=0.325 \mathrm{~kg}\) straight up into the air. In this process, his arm does a total amount of work \(W_{\mathrm{net}}=115 \mathrm{~J}\) on the rock. Calculate the maximum distance, \(h\), above the man's throwing hand that the rock will travel. Neglect air resistance.

Short answer

Question: Calculate the maximum height (h) the rock will travel upwards, given its mass (m) is 0.325 kg, and the initial total amount of work (Wnet) applied to it is 115 J. Assume there's no air resistance. Answer: To find the maximum height (h) the rock will travel upwards, first calculate the initial velocity (v) of the rock using the work-energy theorem. Next, use the kinematic equations with the calculated initial velocity and acceleration due to gravity (9.81 m/s^2) to find the height (h).

Step-by-step solution

1. Calculate the initial kinetic energy and velocity of the rock

To solve this problem, we will first determine the initial kinetic energy of the rock from the work done by the man's arm. According to the work-energy theorem, the net work done on the rock (\(W_{net}\)) equals to the change in its kinetic energy. The rock starts from rest, therefore, its initial kinetic energy is \(0\;\mathrm{J}\). So, we can write: $$W_{\mathrm{net}} = \Delta K.E = K.E_{final} - K.E_{initial}$$ $$115\;\mathrm{J} = \frac{1}{2} m v^2 - 0$$ Now, we can solve for the initial velocity \(v\): $$v = \sqrt{\frac{2 \times 115\;\mathrm{J}}{0.325\;\mathrm{kg}}}$$

2. Use the kinematic equations to calculate the maximum height

Once the initial velocity (\(v\)) is found, we can use one of the kinematic equations that involve height (\(h\)). Since the rock will reach its maximum height when the final velocity is zero, we choose the following equation: $$V_f^2 - V_i^2 = 2 \times g \times h$$ Where: \(V_f\) is the final velocity, equal to \(0\;\mathrm{m/s}\) \(V_i\) is the initial velocity, calculated in step 1 \(g\) is the acceleration due to gravity, which is approximately \(9.81\;\mathrm{m/s^2}\) \(h\) is the height we want to calculate Now, plug in the values and solve for \(h\): $$0 - V_i^2 = 2 \times (-9.81\;\mathrm{m/s^2}) \times h$$ $$h = \frac{V_i^2}{2 \times 9.81\;\mathrm{m/s^2}}$$

Key concepts

Kinetic Energy

In physics, the concept of kinetic energy is fundamental when understanding the movement of objects. It is defined as the energy a body possesses due to its motion. The mathematical expression for kinetic energy (\textbf{K.E}) is \[ K.E = \frac{1}{2} m v^2 \
where \( m \) represents the mass of the object and \( v \) is the velocity at which the object is moving. In the case of the rock thrown into the air by the man, the kinetic energy is the result of the work done by the man's arm. This work transferred energy to the rock, giving it motion and thus kinetic energy.\
\
In every energy transition, like the throw of a rock, some work (\textbf{W}) is done, which is equal to the change in kinetic energy (\textbf{\Delta K.E}). When the object starts from rest, as in our exercise, the initial kinetic energy is zero, and the work done is used to calculate the initial kinetic energy of the rock. Understanding that the rock's initial velocity results from this transfer of energy allows us to calculate how it will move through the air under the force of gravity.

Kinematic Equations

Kinematic equations are a set of four equations that describe the motion of objects without considering the causes of motion (i.e., forces). These equations link displacement, initial velocity, final velocity, acceleration, and time in various combinations to predict an object's motion under constant acceleration.\
\
However, when solving for the maximum height the rock achieves, we look to one specific kinematic equation that relates initial velocity (\textbf{V_i}), final velocity (\textbf{V_f}), acceleration (\textbf{a}), and displacement (\textbf{h}):\
\

\
• \( V_f^2 = V_i^2 + 2 \times a \times h \) \
  \
• When the rock reaches its maximum height, its velocity will be \( V_f = 0 \) since at that instant, it stops rising before beginning to fall back down.\
  \
• As gravity is the only force acting on the rock (neglecting air resistance), the acceleration in the equation is negative gravity (\( -g \) due to the deceleration effect as the rock moves upward).\
  \
• The kinematic equations are an essential tool in physics, enabling us to predict the rock's motion after the energy transfer from throwing.\
  \

By rearranging the equation and plugging in the initial velocity and the value of gravity, we can find the maximum height the rock reaches--a direct application of the kinematic equations that provides students with a deeper understanding of motion under gravity's influence.

Acceleration Due to Gravity

The acceleration due to gravity is a critical concept in physics, impacting the motion of any object near the Earth's surface. Represented by (\textbf{g}), its standard value is \( 9.81\text{ m/s}^2 \) on the surface of the Earth. This acceleration is the rate at which an object's velocity increases as it falls freely towards the Earth's surface, and it operates in the direction of the Earth's center.\
\
When an object is thrown upwards, like the rock in our exercise, it will slow down due to this gravitational pull. Eventually, it will stop briefly at its peak height before falling back down, which corresponds to the final velocity being zero (\textbf{V_f}). This acceleration is constant, and it is always acting downwards, hence the negative sign in the kinematic equation used to calculate the maximum height the rock can reach.\
\
Understanding how gravity influences motion allows us to predict the trajectory of projectiles accurately and is essential in solving a wide range of problems in physics related to motion, energy, and forces.