What is the gravitational potential energy of the obectj

[pola]

hi, please help me figure these problems out. much appreciated.

1. an object of mass .7kg is thrown vertically upwards at a speed of 29 m/s. What is the gravitational potential energy of the obect at the highest point of its trajectory?

2. an object of mass 8.9kg moves with initial speed 21 m/s then interacts with its environment, gaining 1.6J of energy. What is the speed of the object after after the interaction is completed?

 

[pixel01]

1. You have to consider the gravitational potential at the start point as zero. Then the potential at the highest point equals to the dynamic energy of the object or:

E = 1/2mv^2.

2. After the interaction, the dynamic energy is reduced by 1.6J. You can calculate the speed reduction via equ. deltaE=1/2m(deltav)^2. Then the final speed is known by a subtraction

 

[ogb p]

1. The gravitational potential energy of an object is the energy it possesses due to its position in a gravitational field. At the highest point of its trajectory, the object has reached its maximum height and therefore has the maximum potential energy. In order to calculate the gravitational potential energy, we need to know the mass of the object, the acceleration due to gravity (9.8 m/s^2), and the height at the highest point.

Given:
Mass (m) = 0.7 kg
Initial velocity (v) = 29 m/s
Acceleration due to gravity (g) = 9.8 m/s^2
Height at highest point (h) = ?

Using the equation for gravitational potential energy, we can calculate the potential energy at the highest point:

Gravitational potential energy (PE) = mgh
= (0.7 kg)(9.8 m/s^2)(h)
= 6.86 h J

Therefore, the gravitational potential energy of the object at the highest point of its trajectory is 6.86 h J.

2. In this problem, the object starts with an initial speed of 21 m/s and gains 1.6J of energy after interacting with its environment. We can use the conservation of energy principle to determine the final speed of the object.

According to the conservation of energy, the total energy of the system (object + environment) remains constant. Therefore, we can equate the initial kinetic energy of the object to the final kinetic energy after the interaction:

Initial kinetic energy (KEi) = Final kinetic energy (KEf)
= (1/2)mv^2 = (1/2)(8.9 kg)(21 m/s)^2
= 1995.45 J

Since the object gained 1.6 J of energy, the final kinetic energy will be the initial kinetic energy plus the gained energy:

KEf = KEi + Gained energy
= 1995.45 J + 1.6 J
= 1997.05 J

Now, we can rearrange the equation for kinetic energy to solve for the final velocity:

KEf = (1/2)mv^2
v^2 = (2KEf)/m
= (2)(1997.05 J)/(8.9 kg)
= 448.09 m^2/s^2

Taking the square root of both sides, we get:

v