A particle of mass m is initially at rest at | vedclass

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

Question

(B) Given $\frac{dK}{dt} = \gamma t$. Since $K = \frac{1}{2}mv^2$,we have $\frac{dK}{dt} = mv \frac{dv}{dt} = \gamma t$.Thus,$v \frac{dv}{dt} = \frac{

Vedclass · Accepted Answer

$(A), (B), (D)$

Vedclass · Answer

(B) Given $\frac{dK}{dt} = \gamma t$. Since $K = \frac{1}{2}mv^2$,we have $\frac{dK}{dt} = mv \frac{dv}{dt} = \gamma t$.Thus,$v \frac{dv}{dt} = \frac{\gamma t}{m}$.Integrating both sides,$\int v dv = \int \frac{\gamma}{m} t dt$,which gives $\frac{v^2}{2} = \frac{\gamma t^2}{2m}$.Therefore,$v = \sqrt{\frac{\gamma}{m}} t$. This shows that speed is proportional to time,so $(B)$ is true.The acceleration $a = \frac{dv}{dt} = \sqrt{\frac{\gamma}{m}}$,which is a constant. Since $F = ma$,the force $F = \sqrt{\gamma m}$ is also constant,so $(A)$ is true.Since $v = \frac{dx}{dt} = \sqrt{\frac{\gamma}{m}} t$,integrating gives $x = \int \sqrt{\frac{\gamma}{m}} t dt = \frac{1}{2} \sqrt{\frac{\gamma}{m}} t^2$. The distance increases quadratically with time,not linearly,so $(C)$ is false.Since the force is constant,it is a conservative force,so $(D)$ is true.Thus,statements $(A), (B),$ and $(D)$ are true.

Similar Questions

$A$ person of mass $60 \, kg$ wants to lose $5 \, kg$ by going up and down a $10 \, m$ high stairs. Assume he burns twice as much fat while going up than coming down. If $1 \, kg$ of fat is burnt on expending $7000 \, kcal$,how many times must he go up and down to reduce his weight by $5 \, kg$?

$A$ ball is released from a height of $10 \, m$. If after the impact there is a loss of $40 \%$ in its energy,the ball shall rise up to ................. $m$.

$A$ box of mass $1 \ kg$ is placed on a horizontal surface of length $1 \ m$. $A$ force of $8 \ N$ is applied to move it horizontally by $1 \ m$,while simultaneously,its vertical height is increased by $2 \ m$. What is the total work done in Joules? (Take $g = 10 \ m/s^2$)

The bob of a pendulum is released from a horizontal position. If the length of the pendulum is $1.5 \; m$,what is the speed (in $m/s$) with which the bob arrives at the lowermost point,given that it dissipated $5 \%$ of its initial energy against air resistance?

Vedclass Test Series

Exam Paper Generator

Online Exam Module