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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

Expert verified
The work done on the crate is approximately 245.25 J.

Step by step solution

01

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.
02

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).
03

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).
04

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 \( \Delta v_y \approx -8.14\, \text{m/s} \).
05

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, \( \Delta KE = \frac{1}{2} \times 30.0\, \text{kg} \times (31.57 - 15.22)\, \text{m}^2/\text{s}^2 \approx 245.25\, \text{J} \).
06

Determine the Work Done

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

You are applying a constant horizontal force \(\overrightarrow{F} = (-8.00\mathrm{N})\hat{\imath} + (3.00\mathrm{N})\hat{\jmath}\) to a crate that is sliding on a factory floor. At the instant that the velocity of the crate is \(\overrightarrow{\upsilon} = (3.20\mathrm{m/s})\hat{\imath} + (2.20 \mathrm{m/s})\hat{\jmath}\), what is the instantaneous power supplied by this force?

Use the work\(-\)energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resistance in all cases. (a) A branch falls from the top of a 95.0-m-tall redwood tree, starting from rest. How fast is it moving when it reaches the ground? (b) A volcano ejects a boulder directly upward 525 m into the air. How fast was the boulder moving just as it left the volcano?

A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point \(P\), it encounters a rough section. The coefficient of friction there is not constant; it starts at 0.100 at \(P\) and increases linearly with distance past \(P\), reaching a value of 0.600 at 12.5 m past point \(P\). (a) Use the work\(-\)energy theorem to find how far this box slides before stopping. (b) What is the coefficient of friction at the stopping point? (c) How far would the box have slid if the friction coefficient didn't increase but instead had the constant value of 0.100?

A ski tow operates on a 15.0\(^\circ\) slope of length 300 m. The rope moves at 12.0 km/h and provides power for 50 riders at one time, with an average mass per rider of 70.0 kg. Estimate the power required to operate the tow.

A spring of force constant 300.0 N/m and unstretched length 0.240 m is stretched by two forces, pulling in opposite directions at opposite ends of the spring, that increase to 15.0 N. How long will the spring now be, and how much work was required to stretch it that distance?

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