Problem 30 - University Physics with Modern Physics

Question

A 30.0 -kg crate is initially moving with a velocity that has magnitude 3.90 \(\mathrm{m} / \mathrm{s}\) in a direction \(37.0^{\circ}\) west of north. How much work must be done on the crate to change its velocity to 5.62 \(\mathrm{m} / \mathrm{s}\) in a direction \(63.0^{\circ}\) south of east?

Step-by-Step Solution

Answer

The work done is 246.582 J.

Step 1: Find Initial and Final Velocities in Components

First, we need to express the initial and final velocities in terms of their components along the north-south and east-west directions. The initial velocity magnitude is given as 3.90 m/s at an angle of 37.0° west of north. We can find its components using trigonometry:\[ v_{i,north} = 3.90 \cos(37.0°) \]\[ v_{i,west} = 3.90 \sin(37.0°) \]For the final velocity magnitude of 5.62 m/s at 63.0° south of east:\[ v_{f,east} = 5.62 \cos(63.0°) \]\[ v_{f,south} = 5.62 \sin(63.0°) \]

Step 2: Calculate Initial and Final Velocity Vectors

The initial velocity vector can be expressed as:\[ \vec{v}_i = (-v_{i,west}) \hat{i} + (v_{i,north}) \hat{j} \]And the final velocity vector as:\[ \vec{v}_f = (v_{f,east}) \hat{i} + (-v_{f,south}) \hat{j} \]

Step 3: Calculate Change in Velocity Vector

Using the velocity vectors calculated, we find the change in velocity \(\Delta \vec{v}\) by subtracting the initial vector from the final:\[ \Delta \vec{v} = \vec{v}_f - \vec{v}_i \]

Step 4: Use Work-Energy Theorem to Calculate Work Done

According to the work-energy theorem, the work done on an object is equal to the change in kinetic energy. The kinetic energy is given by \(K = \frac{1}{2} m v^2\). We can find the initial and final kinetic energies, \(K_i\) and \(K_f\), and then determine the work done:\[ K_i = \frac{1}{2} \times 30.0 \times (3.90)^2 \]\[ K_f = \frac{1}{2} \times 30.0 \times (5.62)^2 \]Then calculate the work done:\[ W = K_f - K_i \]

Step 5: Substitute and Calculate

Substitute the values for the initial and final kinetic energies and solve for the work done:\[ K_i = 0.5 \times 30 \times 3.90^2 = 228.15 \, \mathrm{J} \]\[ K_f = 0.5 \times 30 \times 5.62^2 = 474.732 \, \mathrm{J} \]The work done is:\[ W = 474.732 - 228.15 = 246.582 \, \mathrm{J} \]

Key Concepts

Understanding Kinetic Energy

Kinetic energy is a form of energy associated with the motion of an object. It's calculated based on the object's mass and velocity, with the formula for kinetic energy given by: \[ K = \frac{1}{2} m v^2 \] This equation tells us that kinetic energy depends on two things:

• The mass (\( m \)) of the object, which in this case is 30.0 kg.
• The square of its velocity (\( v \)), which means even small changes in speed can lead to significant changes in kinetic energy.

In our exercise, the crate's initial kinetic energy was found using its initial velocity of 3.90 m/s. Similarly, the final kinetic energy was determined with a final velocity of 5.62 m/s. The work-energy theorem tells us that the change in kinetic energy is equal to the work done on it. This is crucial because it allows us to understand how much work is needed to change the object's speed and direction.

Velocity Components Explained

When dealing with velocity in physics, it's common to break it down into directional components. This involves splitting a velocity vector into parts aligned with coordinate axes, typically north-south or east-west. It's a vital step because many physics problems, like our crate example, involve multiple directions. To find these components:

• Use the cosine function for the component in the direction of the angle.
• Use the sine function for the component perpendicular to the direction of the angle.

For the crate's initial velocity of 3.90 m/s at an angle 37° west of north:

• North component: \( v_{i,north} = 3.90 \cos(37°) \)
• West component: \( v_{i,west} = 3.90 \sin(37°) \)

Similarly, for the final velocity of 5.62 m/s at 63° south of east:

• East component: \( v_{f,east} = 5.62 \cos(63°) \)
• South component: \( v_{f,south} = 5.62 \sin(63°) \)

By determining these components, you find out how much velocity is directed along or against each axis, simplifying the math for further calculations.

Grasping Change in Velocity

Change in velocity is central to understanding motion in physics. It involves comparing the object's initial and final velocities, seeing how they differ. This difference is not only about speed but also direction. In vector mathematics, change in velocity, denoted as \( \Delta \vec{v} \), is calculated by subtracting one vector from another: \[ \Delta \vec{v} = \vec{v}_f - \vec{v}_i \] This needs us to consider both the size and direction of the velocity vectors, highlighting how they're manipulated to achieve desired changes in motion. For our crate:

• Initial velocity vector \( \vec{v}_i \) points somewhat west of north, represented as \((-v_{i,west}) \hat{i} + (v_{i,north}) \hat{j}\).
• Final velocity vector \( \vec{v}_f \) points south of east, expressed as \((v_{f,east}) \hat{i} + (-v_{f,south}) \hat{j}\).

Finding \( \Delta \vec{v} \) lets us know the precise change needed to move from the initial to the final state, which is crucial for determining the work done using the work-energy theorem. This calculation is foundational in understanding how forces alter an object's motion.