Write the equation of total mechanical energy of a body having mass $m$ and stationary at height $H$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The total mechanical energy $(E)$ of a body is the sum of its kinetic energy $(K)$ and potential energy $(U)$.
$E = K + U$
For a body of mass $m$ at height $H$ that is stationary,its velocity $(v)$ is $0$.
Therefore,the kinetic energy $K = \frac{1}{2}mv^2 = \frac{1}{2}m(0)^2 = 0$.
The gravitational potential energy $U$ at height $H$ relative to the ground is given by $U = mgH$.
Thus,the total mechanical energy is $E = 0 + mgH = mgH$.

Explore More

Similar Questions

$A$ uniform chain of length $L$ and mass $M$ is placed on a smooth table such that one-fourth of its length hangs over the edge. Find the work done to pull the hanging part of the chain onto the table.

$A$ stick of mass $m$ and length $l$ is pivoted at one end and is displaced through an angle $\theta$. The increase in potential energy is ............

$A$ particle is released from rest at the point $x = a$ and moves along the $x$ axis subject to the potential energy function $U(x)$ shown. The particle:

$A$ uniform chain of length $l$ and mass $m$ lies on the surface of a smooth hemisphere of radius $R$ $(R > l)$ with one end tied to the top of the hemisphere as shown in the figure. Gravitational potential energy of the chain with respect to the base of the hemisphere is

The potential energy of a body is given by $U = A - Bx^2$ (where $x$ is the displacement). The magnitude of the force acting on the particle is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo