Question

Draw a new gravitational potential energy vs. height graph to represent the gravitational potential energy if the ball had a mass of 2.00kg. The graph for a 1.00kg ball with an arbitrary initial velocity is provided again as a reference.

Take g = 10.0 m/s² as the acceleration due to gravity.

Answer 1

Concepts and reason

The concept of gravitational potential energy is required to solve the problem. Definition of the gravitational potential energy can determine the gravitational potential energy of a body at various heights. The graphical representation can be done using one axis as the Body's height from the base and one axis as the potential energy. Find out the potential energy at the different at the various location of the body. Draw the table for the potential energy and height. Use the table to draw the graph.

Fundamentals

Potential energy: The body's potential energy is the energy that remains in it due to its position relative to some datum. Gravitational potential energy: Most common type of potential energy that comes across in daily life is gravitational potential energy, which remains in a body due to its position at a height from the ground or any other datum. The body's gravitational potential energy is the energy of the body, which is due to its position at a height from the ground. It can be given as the following, \(P=m g h\)

Where \(P\) is the potential energy of the body, \(m\) is a mass of the body, \(h\) is the distance of the body from the datum, \(g\) is the gravitational acceleration. Graphically the potential energy and height can be represented using ordinate as the potential energy and the abscissa as the height.

Potential energy of the ball is given by, \(P=m g h\)

Substitute \(2 \mathrm{~kg}\) for \(m, 10 \mathrm{~m} / \mathrm{s}^{2}\) for \(g, 2 \cdot 5 \mathrm{~m}\) for \(h\) in equation (1) and calculate the potential energy of the ball at the height of \(2.5 \mathrm{~m}\)

$$ \begin{aligned} P &=(2 \mathrm{~kg})\left(10 \mathrm{~m} / \mathrm{s}^{2}\right)(2 \cdot 5 \mathrm{~m}) \\ &=50 \mathrm{~J} \end{aligned} $$

Substitute \(2 \mathrm{~kg}\) for \(m, 10 \mathrm{~m} / \mathrm{s}^{2}\) for \(g\), \(5 \mathrm{~m}\) for \(h\) in equation (1) and calculate the potential energy of the ball at the height of \(5.0 \mathrm{~m}\)

$$ P=(2 \mathrm{~kg})\left(10 \mathrm{~m} / \mathrm{s}^{2}\right)(5 \mathrm{~m}) $$

\(=100 \mathrm{~J}\)

Substitute \(2 \mathrm{~kg}\) for \(m, 10 \mathrm{~m} / \mathrm{s}^{2}\) for \(g, 7 \cdot 5 \mathrm{~m}\) for \(h\) in equation (1) and calculate the potential energy of the ball at the height of \(7.5 \mathrm{~m}\)

\(P=(2 \mathrm{~kg})\left(10 \mathrm{~m} / \mathrm{s}^{2}\right)(7 \cdot 5 \mathrm{~m})\)

\(=150 \mathrm{~J}\)

Substitute \(2 \mathrm{~kg}\) for \(m, 10 \mathrm{~m} / \mathrm{s}^{2}\) for \(g, 10 \mathrm{~m}\) for \(h\) in equation (1) and calculate the potential energy of the ball at the height of \(10.0 \mathrm{~m}\).

\(P=(2 \mathrm{~kg})\left(10 \mathrm{~m} / \mathrm{s}^{2}\right)(10 \mathrm{~m})\)

$$ =200 \mathrm{~J} $$

Substitute \(2 \mathrm{~kg}\) for \(m, 10 \mathrm{~m} / \mathrm{s}^{2}\) for \(g, 12 \cdot 5 \mathrm{~m}\) for \(h\) in equation (1) and calculate the potential energy of the ball at the height of \(12.5 \mathrm{~m}\).

\(P=(2 \mathrm{~kg})\left(10 \mathrm{~m} / \mathrm{s}^{2}\right)(12 \cdot 5 \mathrm{~m})\)

$$ =250 \mathrm{~J} $$

Substitute \(2 \mathrm{~kg}\) for \(m, 10 \mathrm{~m} / \mathrm{s}^{2}\) for \(g, 15 \mathrm{~m}\) for \(h\) in equation (1) and calculate the potential energy of the ball at the height of \(15.0 \mathrm{~m}\)

\(P=(2 \mathrm{~kg})\left(10 \mathrm{~m} / \mathrm{s}^{2}\right)(15 \mathrm{~m})\)

\(=300 \mathrm{~J}\)

In the above step, potential energy has been calculated for the various position, obtained data can be used to draw potential energy versus height of the ball.

The plot has been drawn using the table, which is obtained by using the potential energy values for the particular height of the ball. The curve obtained is a straight line, which is obvious from the formula of the potential energy as there is only one variable that is the ball's height.