A particle is released from a height H. At a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The total mechanical energy of the particle is conserved,so at any height $h$ from the ground,the sum of potential energy $(PE)$ and kinetic energ

Vedclass · Accepted Answer

$\frac{2 H}{3}, \sqrt{\frac{2 g H}{3}}$

Vedclass · Answer

(D) The total mechanical energy of the particle is conserved,so at any height $h$ from the ground,the sum of potential energy $(PE)$ and kinetic energy $(KE)$ is equal to the initial potential energy at height $H$:$PE + KE = m g H$ ... $(i)$Given that at a certain height,the kinetic energy is half of its potential energy:$KE = \frac{1}{2} PE \implies PE = 2 KE$Substituting this into equation $(i)$:$2 KE + KE = m g H$$3 KE = m g H$$KE = \frac{m g H}{3}$Since $PE = m g h$,we have $PE = 2 KE = 2 \left( \frac{m g H}{3} \right) = \frac{2}{3} m g H$.Equating $m g h = \frac{2}{3} m g H$,we get the height $h = \frac{2 H}{3}$.The speed $v$ of the particle at this height can be found using the work-energy theorem or kinematics. Since the particle falls from rest from height $H$,its speed at height $h$ is given by $v = \sqrt{2 g (H - h)}$.Substituting $h = \frac{2 H}{3}$:$v = \sqrt{2 g (H - \frac{2 H}{3})} = \sqrt{2 g (\frac{H}{3})} = \sqrt{\frac{2 g H}{3}}$.Thus,the height is $\frac{2 H}{3}$ and the speed is $\sqrt{\frac{2 g H}{3}}$.

$A$ particle is released from a height $H$. At a certain height,its kinetic energy is half of its potential energy with reference to the surface of the earth. The height and speed of the particle at that instant are respectively:

Similar Questions

The system shown in the figure is released from rest. The pulley and spring are massless,and friction is absent everywhere. The speed of the $5 \, kg$ block when the $2 \, kg$ block leaves contact with the ground is (take the force constant of the spring $k = 40 \, N/m$ and $g = 10 \, m/s^2$):

$A$ bullet of mass $0.04\; kg$ moving with a speed of $90\; m/s$ enters a heavy wooden block and is stopped after a distance of $60\; cm$. What is the average resistive force exerted by the block on the bullet (in $; N$)?

$A$ particle of mass $500 \,g$ is moving in a straight line with velocity $v = b x^{5/2}$. The work done by the net force during its displacement from $x = 0$ to $x = 4 \,m$ is ................... $J$ (Take $b = 0.25 \,m^{-3/2} s^{-1}$)

$A$ uniform chain of mass $M$ and length $L$ is released from rest from a smooth horizontal surface as shown in the figure. The velocity of the chain at the instant it completely comes off the table will be:

An open knife edge of mass $50 \ g$ is dropped from a height $2 \ m$ on a wooden floor. If the blade penetrates up to a depth of $10 \ cm$ into the wood,the average resistance offered by the wood to the knife edge is $....N$

Vedclass Test Series

Exam Paper Generator

Online Exam Module