The velocity (v) of a particle (under a force | vedclass

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

Question

(C) Given,$v = \frac{k}{\sqrt{x}} = k x^{-1/2}$,where $k$ is a constant.Acceleration $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{

Vedclass · Accepted Answer

$F \propto \frac{1}{x^2}$

Vedclass · Answer

(C) Given,$v = \frac{k}{\sqrt{x}} = k x^{-1/2}$,where $k$ is a constant.Acceleration $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$.First,find $\frac{dv}{dx}$:$\frac{dv}{dx} = k \cdot (-\frac{1}{2}) x^{-3/2} = -\frac{k}{2x^{3/2}}$.Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula:$a = (k x^{-1/2}) \cdot (-\frac{k}{2x^{3/2}}) = -\frac{k^2}{2x^2}$.According to Newton's second law,$F = ma$.The magnitude of the force is $F = |ma| = m \cdot \frac{k^2}{2x^2}$.Since $m$ and $k$ are constants,$F \propto \frac{1}{x^2}$.

The velocity $(v)$ of a particle (under a force $F$) depends on its distance $(x)$ from the origin (with $x > 0$) as $v \propto \frac{1}{\sqrt{x}}$. Find how the magnitude of the force $(F)$ on the particle depends on $x$.

Similar Questions

$A$ constant force acting on a body of mass $3.0 \, kg$ changes its speed from $2.0 \, m/s$ to $3.5 \, m/s$ in $25 \, s$. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

The force required to stop a body of mass $10 \,kg$ moving along a straight line path with a velocity of $10 \,ms^{-1}$ in a time of $10 \,s$ is (in $\,N$)

$A$ cricket player catches a ball of mass $120 \,g$ moving with $25 \,m/s$ speed. If the catching process is completed in $0.1 \,s$, then the magnitude of force exerted by the ball on the hand of the player will be (in $SI$ unit):

$A$ car of mass $1500 \,kg$ is moving with $20 \,ms^{-1}$ velocity. If the brakes are applied,it comes to rest in $5 \,s$. The retarding force is: (in $\,N$)

$A$ force $\vec{F}=(40 \hat{i}+10 \hat{j}) \, N$ acts on a body of mass $5 \, kg$. If the body starts from rest,its position vector $\vec{r}$ at time $t=10 \, s$ will be -

Vedclass Test Series

Exam Paper Generator

Online Exam Module