At time t=0 s particle starts moving along the x-axis. If its kinetic energy increases uniformly with time t, the net force acting on it must be proportional to

Question

At time t=0 s particle starts moving along the x-axis. If its kinetic energy increases uniformly with time t, the net force acting on it must be proportional to (A) √t (B) constant (C) t (D) (1/√t)

Tardigrade Admin · Accepted Answer

Given, (d k/d t)= constant ⇒ k propto t ⇒ v propto √t Also, P=F v=(d k/d t)= constant ⇒ F propto (1/v) ⇒ F propto (1/√t)

Q. At time $t = 0$ s particle starts moving along the $x$ -axis. If its kinetic energy increases uniformly with time $t$ , the net force acting on it must be proportional to

$t$

constant

$t$

$\frac{1}{t}$

Given, $\frac{d k}{d t} =$ constant
$\Rightarrow k \propto t \Rightarrow v \propto t$
Also, $P = F v = \frac{d k}{d t} =$ constant
$\Rightarrow F \propto \frac{1}{v} \Rightarrow F \propto \frac{1}{t}$

Q. At time t=0 s particle starts moving along the x-axis. If its kinetic energy increases uniformly with time t, the net force acting on it must be proportional to

t​

constant

t

t​1​

Given, dtdk​= constant ⇒k∝t⇒v∝t​ Also, P=Fv=dtdk​= constant ⇒F∝v1​⇒F∝t​1​

Q. At time $t = 0$ s particle starts moving along the $x$ -axis. If its kinetic energy increases uniformly with time $t$ , the net force acting on it must be proportional to

$t$

$t$

$\frac{1}{t}$

Given, $\frac{d k}{d t} =$ constant
$\Rightarrow k \propto t \Rightarrow v \propto t$
Also, $P = F v = \frac{d k}{d t} =$ constant
$\Rightarrow F \propto \frac{1}{v} \Rightarrow F \propto \frac{1}{t}$