A body of mass m starts moving from rest alon | vedclass

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

Question

(A) Given the velocity $v = a\sqrt{s}$.Since $v = \frac{ds}{dt}$,we have $\frac{ds}{dt} = a\sqrt{s}$.Separating variables,we get $\int s^{-1/2} ds = \

Vedclass · Accepted Answer

$\frac{1}{8} m a^4 t^2$

Vedclass · Answer

(A) Given the velocity $v = a\sqrt{s}$.Since $v = \frac{ds}{dt}$,we have $\frac{ds}{dt} = a\sqrt{s}$.Separating variables,we get $\int s^{-1/2} ds = \int a dt$.Integrating both sides,$2\sqrt{s} = at + C$. Since the body starts from rest at $t=0$,$s=0$,so $C=0$.Thus,$\sqrt{s} = \frac{at}{2}$,which implies $s = \frac{a^2 t^2}{4}$.Differentiating $s$ with respect to $t$,we get velocity $v = \frac{ds}{dt} = \frac{a^2 t}{2}$.Differentiating $v$ with respect to $t$,we get acceleration $a_{acc} = \frac{dv}{dt} = \frac{a^2}{2}$.According to the work-energy theorem,the total work done is equal to the change in kinetic energy: $W = \Delta K = \frac{1}{2} m v^2$.Substituting $v = \frac{a^2 t}{2}$,we get $W = \frac{1}{2} m \left( \frac{a^2 t}{2} \right)^2 = \frac{1}{2} m \left( \frac{a^4 t^2}{4} \right) = \frac{1}{8} m a^4 t^2$.

$A$ body of mass $m$ starts moving from rest along the $x-$axis such that its velocity varies as $v = a\sqrt{s}$,where $a$ is a constant and $s$ is the distance covered by the body. The total work done by all the forces acting on the body in the first $t$ seconds after the start of the motion is:

Similar Questions

$A$ body of mass $0.5\, kg$ travels on a straight-line path with velocity $v = (3x^2 + 4)\, m/s$. The net work done by the force during its displacement from $x = 0$ to $x = 2\, m$ is $.......J$.

$A$ body of mass $m$ is moving with uniform acceleration such that at time $t_1$ its velocity is $v$. What is the work done on the body at any time $t$?

Under the action of a force, a $2 \,kg$ body moves such that its position $x$ as a function of time $t$ is given by $x = \alpha t^2 / 2$, where $x$ is in meters, $t$ is in seconds, and $\alpha = 1 \,m/s^2$. The work done by the force in the first two seconds is: (in $\,J$)

$A$ block of mass $100\, g$ slides on a rough horizontal surface. If the speed of the block decreases from $10\, m/s$ to $5\, m/s,$ the thermal energy developed in the process is............. $J$.

The potential energy of a $1 \ kg$ particle free to move along the $x-$axis is given by: $U(x) = (\frac{x^4}{4} - \frac{x^2}{2}) \ J$. The total mechanical energy of the particle is $2 \ J$. Then,the maximum speed (in $m/s$) is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module