Question

14. Show that when a body is dropped from a certain height, the sum of its kinetic energy
at any instant during its fall is constant. (5)
15. Briefly describing the gravitational potential energy, deduce an expression for the
gravitational potential energy of a body of mass m placed at a height h, above the
ground. ​

Answer 1

Answer:

14.)Show that when a body is dropped from a certain height, the sum of its kinetic energy and potential energy at any instant during its fall is constant.

→ Conservation of energy during free fall. Thus, throughout its motion the total energy of the body is constant and is equal to mgh.

E(k) = m×g×h

15.)Near the surface of the Earth, the work done in lifting an object through a height h is the product mgh , so U=mgh .

The gravitational potential energy, U , of a system of masses m1 and M2 at a distance r using gravitational constant G is

U = −Gm1M2r

U = − G m 1 M 2 r .

Explanation:

Answer 2

Answer:

14. When a body is dropped from a certain height, the sum of its kinetic energy at any instant during its fall is constant it is proved

Potential energy is equivalent (in size, however negative) to the work done by the gravitational field moving a body to its given situation in space from boundlessness. The expression of potential energy $E_{\{p\}}=m g h$

Explanation:

14.

Tip:

• Conservation of energy during the free fall of a body. Let a body of mass $m$ lying at position A be moved to a position $B$ through a height $h$ (Fig). The work done on the body is mgh. Since the body is at rest at position B, therefore,

Explanation:

• K.E. at $B=0$
• P.E. at $B=m g h$
• Total energy at $B=m g h+0=m g h$
• If the body falls from $B to C$ through a distance $x$, the body possesses both P.E and K.E.
• Now P.E. at $C=m g(h-x)$.
• Let $v$ be the velocity acquired by the body when it reaches $C$.
• $\therefore \quad v^{2}-0=2 g x or v^{2}=2 g x$
• $\therefore \quad K.E. at C=\frac{1}{2} m v^{2}=\frac{1}{2} m \times 2 g x=m g x .$
• So total energy at $C=m g(h-x)+m g x=m g h$
• When the body reaches the initial position $A$
• P.E. at $A=0$
• Let $V$ be the velocity of the body at $A$. $V$ is given by
• $V^{2}-0=2 g h \quad \text { or } \quad V^{2}=2 g h$
• $\therefore \quad K.E. at A=\frac{1}{2} m V^{2}=\frac{1}{2} m \times 2 g h=m g h$
• $\therefore Total energy at A=m g h$

Final answer:

Thus, throughout its motion the total energy of the body is constant and is equal to mgh. This verifies the law of conservation of energy.

15. Tip

• The gravitational potential (V) at an area is the gravitational possible energy (U) at that area per unit mass: where m is the mass of the article.
• Potential energy is equivalent (in size, however negative) to the work done by the gravitational field moving a body to its given situation in space from boundlessness.

Explanation:

• Think about a group of mass m lying at point An on the world's surface, where its potential energy is taken as zero. Its weight mg acts vertically downwards. To lift the body to another position B at a tallness h, we need to apply a base power equivalent to mg the upward way. So work is done on the body against the power of gravity.
• Work done $= Force x Displacement$
• $W=F s$
• $F=m g$ $s=h \\ \therefore W=m g \times h=m g h$ The work done (mgh) on the body is equivalent to the increase in the energy of the body. This is the expected energy of the Body. Thus, $E_{\{p\}}=m g h$

Final answer

Potential energy is equivalent (in size, however negative) to the work done by the gravitational field moving a body to its given situation in space from boundlessness.

The expression of potential energy $E_{\{p\}}=m g h$