Question

Show that the total mechanical energy for a freely falling body above the height .
Height (x) in constant​

Answer 1

Question:

Show that the total mechanical energy for a freely falling body above the height is conserved. Height (x) in constant​.

Answer:

• Read explanation.

Explanation:

Mechanical Energy:

• It is the sum of kinetic energy as well as potential energy.
• v = K.E + P.E

• ${\sf{v=\dfrac{1}{2}\;mv^{2}+mgh}}$

Proof:

Ps : To better understand please refers to attachment for diagram.

At point "A",

$\implies{\sf{(K.E)_{A}=\dfrac{1}{2}mv^{2}_{A}=0}}$

$\implies{\sf{(P.E)_{A}=mgh}}$

${\sf{So,\;(M.E)_{A}=(K.E)_{A}+(P.E)_{A}}}$

$\implies{\sf{\;(M.E)_{A}=0+mgh}}$

$\implies{\boxed{\sf{\;(M.E)_{A}=mgh\;\;\;\;..(1)}}}$

At point "B",

$\implies{\sf{(K.E)_{B}=\dfrac{1}{2}mv^{2}_{B}\;\;\;\;..(2)}}$

From, 3rd equation of motion.

$\implies{\sf{v^{2}=u^{2}+2as}}$

$\implies{\sf{v^{2}_{B}=0+2gx\;\;\;\;\;[h=x\;and\;a\;repleaced\;by\;g]}}$

$\implies{\sf{v^{2}_{B}=2gx}}$

Now, Put the value of $\sf{V^{2}_{B}}$ in equation (2).

$\implies{\sf{(K.E)_{B}=\dfrac{1}{2}m(2gx)}}$

$\implies{\sf{(K.E)_{B}=mgx}}$

$\implies{\sf{(P.E)_{B}=mg(h-x)}}$

${\sf{So,\;(M.E)_{B}=(K.E)_{A}+(P.E)_{A}}}$

$\implies{\sf{\;(M.E)_{B}=mgx+mg(h-x)}}$

$\implies{\sf{\;(M.E)_{B}=mgx+mgh-mgx}}$

$\implies{\boxed{\sf{\;(M.E)_{B}=mgh\;\;\;\;..(3)}}}$

At point "C",

$\implies{\sf{(K.E)_{C}=\dfrac{1}{2}mv^{2}_{C}\;\;\;\;..(4)}}$

From, 3rd equation of motion.

$\implies{\sf{V^{2}_{c} =u^{2}+2gs}}$

$\implies{\sf{V^{2}_{c} =0+2gh}}$

Now, put the value of $\sf{v^{2}_{c}}$ in equation (4),

$\implies{\sf{(K.E)_{C}=\dfrac{1}{2}m(2gh)}}$

$\implies{\sf{(K.E)_{C}=mgh}}$

$\implies{\sf{(P.E)_{C}=mg(0)=0}}$

${\sf{So,\;(M.E)_{C}=(K.E)_{A}+(P.E)_{A}}}$

$\implies{\sf{\;(M.E)_{B}=mgh+0}}$

$\implies{\boxed{\sf{\;(M.E)_{C}=mgh\;\;\;\;..(5)}}}$

Hence, From equation (1), (3) and (5) we may conclude that the energy that the object remains conserve during free fall of the object.