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A \(5.75-\mathrm{kg}\) object is initially moving so that its \(x\) -component of velocity is \(6.00 \mathrm{~m} / \mathrm{s}\) and its \(y\) -component of velocity is \(-2.00 \mathrm{~m} / \mathrm{s}\), (a) What is the kinetic energy of the object at this time? (b) Find the change in kinetic energy of the object if its velocity changes so that its new \(x\) -component is \(8.50 \mathrm{~m} / \mathrm{s}\) and its new \(y\) -component is \(5.00 \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
The initial kinetic energy of the object is the result from step 2. The change in kinetic energy of the object is the result from step 5.

Step by step solution

01

Calculate the Magnitude of the Initial Velocity

Use the Pythagorean theorem to calculate the magnitude of the velocity vector \(\sqrt{(6.00 \mathrm{~m/s})^{2} + (-2.00 \mathrm{~m/s})^{2}}\).
02

Compute the Initial Kinetic Energy

Substitute the magnitude of the velocity and the mass of the object in the kinetic energy formula \(K = 1/2 m v^{2}\). Thus \((1/2) (5.75 \mathrm{~kg}) (\mbox{result from step 1})^{2}\).
03

Calculate the Magnitude of the Final Velocity

Once again, apply the Pythagorean theorem, but now use the final velocity components \(\sqrt{(8.50 \mathrm{~m/s})^{2} + (5.00 \mathrm{~m/s})^{2}}\). This gives the magnitude of the final velocity.
04

Compute the Final Kinetic Energy

Using the result from step 3 and the mass of the object, calculate the final kinetic energy using the formula \(K = 1/2 m v^{2}\). So we have \((1/2) (5.75 \mathrm{~kg}) (\mbox{result from step 3})^{2}\).
05

Calculate the Change in Kinetic Energy

Subtract the initial kinetic energy (result from step 2) from the final kinetic energy (result from step 4) to get the change in kinetic energy.

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Key Concepts

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

Pythagorean theorem in velocity calculations
When dealing with velocities in two dimensions, it's important to know the overall speed, or the magnitude of the velocity. This is where the Pythagorean theorem comes in handy. Imagine you have two velocity components: one along the x-axis and another along the y-axis. To find the overall velocity, treat these components as the sides of a right triangle.

According to the Pythagorean theorem, the length of the hypotenuse (resultant velocity) is the square root of the sum of the squares of these components. In our example with initial velocities, you'd calculate this as:
  • Initial x-component = 6.00 m/s
  • Initial y-component = -2.00 m/s
  • Magnitude of velocity = \(\sqrt{(6.00)^2 + (-2.00)^2}\)
So, the magnitude of the velocity gives you a single value representing the object's speed in both directions together.
Understanding velocity components
Velocity, simply put, is an object's speed and direction. In physics, we often break this down into components, especially when working with two or three dimensions. These components are vectors pointing in specific directions. For example, a velocity vector in 2D can be split into two components: one along the x-axis and one along the y-axis.

In our given exercise, the object starts with a velocity that has an x-component of 6.00 m/s and a y-component of -2.00 m/s. The signs indicate direction; positive x means rightwards, and negative y means downwards. When the velocity changes, its components become 8.50 m/s and 5.00 m/s. This indicates a new direction and speed; positive values here mean movement to the right and upwards.

By looking at the components separately, you can easily understand how the object's motion changes in each direction.
Calculating change in kinetic energy
Kinetic energy is the energy that an object possesses due to its motion. It is calculated using the formula \(K = \frac{1}{2} m v^2\), where \(m\) is the mass of the object, and \(v\) is its velocity. In our problem, kinetic energy is calculated before and after a change in velocity.

First, you find the initial kinetic energy using the magnitude of the initial velocity (from the Pythagorean theorem) and the mass, then you do the same for the new, or final, velocity.

Let's break it down:
  • Initial KE: \(\frac{1}{2} \times 5.75 \times (\text{initial velocity magnitude})^2\)
  • Final KE: \(\frac{1}{2} \times 5.75 \times (\text{final velocity magnitude})^2\)
Finally, the change in kinetic energy is simply the difference between the final and initial kinetic energies. This value tells you how much the energy has increased or decreased due to the change in velocity.

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Most popular questions from this chapter

A \(7.80\) -g bullet moving at \(575 \mathrm{~m} / \mathrm{s}\) penetrates a tree trunk to a depth of \(5.50 \mathrm{~cm}\). (a) Use work and energy considerations to find the average frictional force that stops the bullet. (b) Assuming the frictional force is constant, determine how much time elapses between the moment the bullet enters the tree and the moment it stops moving.

A horizontal spring attached to a wall has a force constant of \(850 \mathrm{~N} / \mathrm{m} . \mathrm{A}\) block of mass \(1.00 \mathrm{~kg}\) is attached to the spring and oscillates freely on a horizontal, frictionless surface as in Active Figure \(5.20 .\) The initial goal of this problem is to find the velocity at the equilibrium point after the block is released. (a) What objects constitute the system, and through what forces do they interact? (b) What are the two points of interest? (c) Find the energy stored in the spring when the mass is stretched \(6.00 \mathrm{~cm}\) from equilibrium and again when the mass passes through cquilibrium after being released from rest. (d) Write the conservation of energy equation for this situation and solve it for the speed of the mass as it passes equilibrium. Substitute to obtain a numerical value. (e) What is the speed at the halfway point? Why isn't it half the speed at equilibrium?

A \(25.0\) -kg child on a \(2.00\) -m-long swing is released from rest when the ropes of the swing make an angle of \(30.0^{\circ}\) with the vertical. (a) Neglecting friction, find the child's speed at the lowest position. (b) If the actual speed of the child at the lowest position is \(2.00 \mathrm{~m} / \mathrm{s}\), what is the mechanical energy lost due to friction?

(a) A child slides down a water slide at an amusement park from an initial height \(h\). The slide can be considered frictionless because of the water flowing down it. Can the equation for conservation of mechanical energy be used on the child? (b) Is the mass of the child a factor in determining his speed at the bottom of the slide? (c) The child drops straight down rather than following the curved ramp of the slide. In which case will he be traveling faster at ground level? (d) If friction is present, how would the conservation-of-energy equation be modified? (c) Find the maximum speed of the child when the slide is frictionless if the initial height of the slide is \(12.0 \mathrm{~m}\).

A \(50-\mathrm{kg}\) pole vaulter running at \(10 \mathrm{~m} / \mathrm{s}\) vaults over the bar. Her speed when she is above the bar is \(1.0 \mathrm{~m} / \mathrm{s}\). Neglect air resistance, as well as any energy absorbed by the pole, and determine her altitude as she crosses the bar.

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