SATHEE: Kinematics Question 484

Kinematics Question 484

Question: The kinetic energy $k$ of a particle moving along a circle of radius $R$ depends on the distance covered $s$ as $k = a s^{2}$ where $a$ it’s a constant. The force acting on the particle is [MNR 1992; JIPMER 2001, 02; AMU (Engg.) 1999]

Options:

A) $2 a \frac{s^{2}}{R}$

B) $2 a s {(1 + \frac{s^{2}}{R^{2}})}^{1 / 2}$

C) $2 a s$

D) $2 a \frac{R^{2}}{s}$

Show Answer

Answer:

Correct Answer: B

Solution:

According to given problem $\frac{1}{2} m v^{2} = a s^{2}$

$\Rightarrow v = s \sqrt{\frac{2 a}{m}}$

So $a_{R} = \frac{v^{2}}{R} = \frac{2 a s^{2}}{m R}$ -(i)

Further more as $a_{t} = \frac{d v}{d t} = \frac{d v}{d s} \cdot \frac{d s}{d t} = v \frac{d v}{d s}$ -(ii)

(By chain rule) Which in light of equation (i) i.e. $v = s \sqrt{\frac{2 a}{m}}$

yields $a_{t} = [s \sqrt{\frac{2 a}{m}}] [\sqrt{\frac{2 a}{m}}] = \frac{2 a s}{m}$ -(iii)

So that $a = \sqrt{a_{R}^{2} + a_{t}^{2}} = \sqrt{{[\frac{2 a s^{2}}{m R}]}^{2} + {[\frac{2 a s}{m}]}^{2}}$

Hence $a = \frac{2 a s}{m} \sqrt{1 + {[s / R]}^{2}}$

$F = m a = 2 a s \sqrt{1 + {[s / R]}^{2}}$

Kinematics Question 484

Question: The kinetic energy $k$ of a particle moving along a circle of radius $R$ depends on the distance covered $s$ as $k = a s^{2}$ where $a$ it’s a constant. The force acting on the particle is [MNR 1992; JIPMER 2001, 02; AMU (Engg.) 1999]

Options:

Answer:

Solution:

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Kinematics Question 484

Question: The kinetic energy k of a particle moving along a circle of radius R depends on the distance covered s as k=as2 where a it’s a constant. The force acting on the particle is [MNR 1992; JIPMER 2001, 02; AMU (Engg.) 1999]

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: The kinetic energy $k$ of a particle moving along a circle of radius $R$ depends on the distance covered $s$ as $k = a s^{2}$ where $a$ it’s a constant. The force acting on the particle is [MNR 1992; JIPMER 2001, 02; AMU (Engg.) 1999]