The kinetic energy K of a particle moving in | vedclass

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

Question

(A) Given,kinetic energy $K = as^2$.We know that $K = \frac{1}{2}mv^2$,where $m$ is the mass and $v$ is the velocity of the particle.So,$\frac{1}{2}mv

Vedclass · Accepted Answer

$2as$

Vedclass · Answer

(A) Given,kinetic energy $K = as^2$.We know that $K = \frac{1}{2}mv^2$,where $m$ is the mass and $v$ is the velocity of the particle.So,$\frac{1}{2}mv^2 = as^2$.Differentiating both sides with respect to time $t$:$\frac{d}{dt}(\frac{1}{2}mv^2) = \frac{d}{dt}(as^2)$$\frac{1}{2}m(2v \frac{dv}{dt}) = a(2s \frac{ds}{dt})$Since $F = m \frac{dv}{dt}$ and $v = \frac{ds}{dt}$,we substitute these into the equation:$mv \frac{dv}{dt} = 2as \frac{ds}{dt}$$F \cdot v = 2as \cdot v$$F = 2as$.

The kinetic energy $K$ of a particle moving in a straight line depends upon the distance $s$ as $K = as^2$. The force acting on the particle is

Similar Questions

If a constant force of $(2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \text{ N}$ acting on a body of mass $5 \text{ kg}$ displaces it from $(3 \hat{i} - 4 \hat{k}) \text{ m}$ to $(2 \hat{i} + 2 \hat{j} + 3 \hat{k}) \text{ m}$, then the work done by the force on the body is (in $\text{ J}$)

$A$ body of mass $m$ accelerates uniformly from rest to a speed $v_0$ in time $t_0$. The work done on the body till any time $t$ is

$A$ force of magnitude $50\, N$ acting along $\hat{i} + \hat{j} + \hat{k}$ displaces a point mass from $(5, 9, 7)$ to $(4, 8, 6)$. The work done by this force during this displacement is:

Write the definition of work and kinetic energy.

$A$ body of mass $m$ kg is lifted by a man to a height of one metre in $30\, s$. Another man lifts the same mass to the same height in $60\, s$. The work done by them are in the ratio:

Vedclass Test Series

Exam Paper Generator

Online Exam Module