Question

A 30.0-kg crate is initially moving with a velocity that has magnitude 3.90 m/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 m/s in a direction 63.0\(^\circ\) south of east?

Short answer

The work done on the crate is approximately 245.25 J.

Step-by-step solution

Identify Initial and Final Velocities

The initial velocity \( \vec{v_i} \) is 3.90 m/s at 37.0° west of north, and the final velocity \( \vec{v_f} \) is 5.62 m/s at 63.0° south of east. These will later be resolved into components for calculation purposes.

Resolve Initial Velocity into Components

Calculate the north (y-axis) and west (x-axis) components of the initial velocity. The components are: \( v_{ix} = 3.90 \sin(37.0^\circ) \) and \( v_{iy} = 3.90 \cos(37.0^\circ) \). Evaluate these values: \( v_{ix} \approx 2.35\, \text{m/s} \) (west), and \( v_{iy} \approx 3.12\, \text{m/s} \) (north).

Resolve Final Velocity into Components

Calculate the south (negative y-axis) and east (positive x-axis) components of the final velocity. The components are: \( v_{fx} = 5.62 \cos(63.0^\circ) \) and \( v_{fy} = 5.62 \sin(63.0^\circ) \). Evaluate these values: \( v_{fx} \approx 2.54\, \text{m/s} \) (east), and \( v_{fy} \approx -5.02\, \text{m/s} \) (south).

Calculate Change in Velocity Components

Find the change in velocity components: \( \Delta v_x = v_{fx} - (-v_{ix}) = 2.54 + 2.35 \) and \( \Delta v_y = v_{fy} - v_{iy} = -5.02 - 3.12 \). Thus, \( \Delta v_x \approx 4.89 \, \text{m/s} \) and \( \Delta v_y \approx -8.14\, \text{m/s} \).

Calculate Change in Kinetic Energy

The change in kinetic energy \( \Delta KE \) is given by \( \Delta KE = \frac{1}{2} m (v_f^2 - v_i^2) \). First, calculate \( v_f^2 = v_{fx}^2 + v_{fy}^2 \approx 2.54^2 + (-5.02)^2 \approx 31.57\, \text{m}^2/\text{s}^2 \) and \( v_i^2 = v_{ix}^2 + v_{iy}^2 \approx 2.35^2 + 3.12^2 \approx 15.22\, \text{m}^2/\text{s}^2 \). Then, \( \Delta KE = \frac{1}{2} \times 30.0\, \text{kg} \times (31.57 - 15.22)\, \text{m}^2/\text{s}^2 \approx 245.25\, \text{J} \).

Determine the Work Done

The work done on the crate is equal to the change in kinetic energy. Therefore, the work done \( W = \Delta KE \approx 245.25\, \text{J} \).

Key concepts

Kinetic Energy

Kinetic energy is a fundamental concept in physics that describes the energy possessed by a body due to its motion. It is calculated using the formula \[ KE = \frac{1}{2} m v^2 \] where \( KE \) represents kinetic energy, \( m \) is mass, and \( v \) is velocity. The velocity is crucial as the energy increases with the square of velocity, meaning even a small increase in speed can result in a significant rise in kinetic energy.

In the exercise, the change in the crate's kinetic energy was calculated to determine the work done to alter its velocity. Initially, the velocity of the crate led to a certain kinetic energy, and the new velocity provided a different energy value. The difference between these two energy values was used to find how much work was needed to achieve this change. This process illustrates the close link between work and kinetic energy, showcasing how work done on an object can change its kinetic energy by altering its velocity.

Velocity Components

Understanding velocity components is essential when dealing with vectors in physics. Velocity is a vector, which means it has both magnitude and direction. When a velocity vector is given, like in our exercise, it's often at an angle, and resolving it into components simplifies the analysis. These components represent movements along the perpendicular axes, typically the x-axis (horizontal) and y-axis (vertical).

For our example exercise, we had to resolve both the initial and final velocities into their respective components. This involved using trigonometric functions: - The sine function was used for calculating the component along the west-east axis. - The cosine function calculated the component along the north-south axis.

This resolution makes it easier to perform calculations such as those required for determining changes in velocity and kinetic energy, as it allows us to work with straight-line components rather than angled vectors. By splitting a velocity into components, we simplify the physics problem substantially.

Vector Resolution

In physics, vector resolution refers to breaking down a vector into its component parts. Any vector, including velocity and force, can be resolved into components. This is especially useful in 2D and 3D motion analyses. By resolving vectors, we can analyze multi-directional movements more easily using orthogonal axes, typically labeled x and y-axes, or even the z-axis for three dimensions.

In the exercise, the process of vector resolution was applied to the velocities given in angled directions. By using trigonometry: - The cosine function helped calculate the aligned component on the horizontal or vertical depending on the vector.- The sine function determined the perpendicular component.

• The x-component is usually \( v_x = v \cos(\theta) \)
• The y-component is usually \( v_y = v \sin(\theta) \)

These components let us precisely find how much of the velocity is aimed along specific directions and facilitate understanding the effect of forces along these pathways. This is crucial when calculating other vector quantities like force and momentum. The resolution of vectors into components is indispensable for systematically solving complex physics problems.