When the batter hit the foul ball, the baseball moved upward to a delighted fan in the top deck.

Physics

1 answer:

oee [108]3 years ago

6 0

Answer:

B) Kinetic energy increases, potential energy decreases

Explanation:

In a given system, when a body is at rest, v =0m/s, the kinetic energy is at zero while the potential energy is at maximum. However, when a body is in motion with a velocity = v, the potential energy is at zero while the kinetic energy is at maximum.

Before this happen, the a body at rest (P.E = max) is set on motion, the kinetic energy gradually increases till it converts all the potential energy in the system to kinetic energy and then reverses back when the body goes to rest again.

In this case, before the batter hits the ball, the kinetic energy was at zero while the potential energy was at maximum. However, when he hits the ball and sets it into motion with a velocity V, the potential energy converts to kinetic energy and moves the ball with that energy till it has expanded it and comes to rest.

Potential Energy → Kinetic Energy → Potential Energy.

That's how the system keeps changing.

You might be interested in

A 2kg mass is initially at rest on a frictionless surface. A 45N force acts at an angle of 300

raketka [301]

Answer:

a) The work done by the force is 136.400 joules.

b) The speed of the mass at the end of the 3.5 meters is approximately 11.679 meters per second.

Explanation:

The correct statement is shown below:

A 2-kg mass is initially at reat on a frictionless surface. A 45 N force acts at an angle of 30º to the horizontal for a distance of 3.5 meters:

a) Find the work done by the force.

b) Find the speed of the mass at the end of the 3.5 meters.

a) Given that a external constant force is acting on the mass on a frictionless surface, the work done by the force ( $W_{F}$ ), measured in joules, is:

$W_{F} = F\cdot \Delta s \cdot \cos \theta$ (1)

Where:

$F$ - External constant force exerted on the mass, measured in newtons.

$\Delta s$ - Horizontal travelled distance, measured in meters.

$\theta$ - Direction of the external force regarding the horizontal, measured in sexagesimal degrees.

If we know that $F = 45\,N$ , $\Delta s = 3.5\,m$ and $\theta = 30^{\circ}$ , then the work done by the force is:

$W_{F} = (45\,N)\cdot (3.5\,m)\cdot \cos 30^{\circ}$

$W_{F} = 136.400\,J$

The work done by the force is 136.400 joules.

b) The final speed of the mass at the end ot the 3.5 meter is calculated by means of the Work-Energy Theorem, which means that the work done on the mass is transformed into a translational kinetic energy. That is:

$W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})$ (2)