A particle of mass $10\,g$ moves along a circle of radius $6.4\,cm$ with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to $8 \times 10^{-4} \, J$ by the end of the second revolution after the beginning of the motion ?

Question

A force of 10 N is applied to move a body of mass 5 kg over a distance of 3 meters. Find the work done by the force.

Answer 1

To find the tangential acceleration, we need to utilize the information provided about kinetic energy and the motion of the particle. Let's solve it step-by-step:

Given that the particle has a mass $m = 10\, \text{g} = 0.01\, \text{kg}$ and moves along a circle of radius $R = 6.4\, \text{cm} = 0.064\, \text{m}$.
We need to find the tangential acceleration $a_t$ when the kinetic energy $KE = 8 \times 10^{-4} \, \text{J}$ after completing two revolutions.
The kinetic energy formula for a rotating object is: $$KE = \frac{1}{2} m v^2$$ where $v$ is the linear velocity.
From the formula: $$8 \times 10^{-4} = \frac{1}{2} \times 0.01 \times v^2$$ Solving for $v^2$, we have: $$v^2 = \frac{2 \times 8 \times 10^{-4}}{0.01} = 0.16$$ Thus, $v = \sqrt{0.16} = 0.4 \, \text{m/s}$.
The particle undergoes tangential acceleration, so we apply the equation for final velocity under constant acceleration: $$v = u + a_t t$$ where $u = 0$ (initial velocity), $v = 0.4 \, \text{m/s}$. The number of revolutions is 2, so the distance traveled is: $$s = 2 \times 2\pi R = 4\pi \times 0.064 \, \text{m} = 0.256\pi \, \text{m}$$
Using the equation for motion: $$s = ut + \frac{1}{2} a_t t^2$$ Simplifying, $$0.256\pi = \frac{1}{2} a_t t^2$$ As $v = a_t t$, $$0.4 = a_t \cdot t \implies t = \frac{0.4}{a_t}$$ Substituting t: $$0.256\pi = \frac{1}{2} a_t \left(\frac{0.4}{a_t}\right)^2 = \frac{0.08}{2a_t}$$ Solve for $a_t$: $$0.256\pi = \frac{0.08}{2a_t}$$ $$a_t = \frac{0.08}{2 \times 0.256\pi}$$ = 0.1 \text{m/s}^2

The correct magnitude of the tangential acceleration a_t is thus $0.1 \, \text{m/s}^2$.

Hence, the correct answer is $0.1 \, \text{m/s}^2$.

Answer 2

To find the tangential acceleration, we need to utilize the information provided about kinetic energy and the motion of the particle. Let's solve it step-by-step:

Given that the particle has a mass $m = 10\, \text{g} = 0.01\, \text{kg}$ and moves along a circle of radius $R = 6.4\, \text{cm} = 0.064\, \text{m}$.
We need to find the tangential acceleration $a_t$ when the kinetic energy $KE = 8 \times 10^{-4} \, \text{J}$ after completing two revolutions.
The kinetic energy formula for a rotating object is: $$KE = \frac{1}{2} m v^2$$ where $v$ is the linear velocity.
From the formula: $$8 \times 10^{-4} = \frac{1}{2} \times 0.01 \times v^2$$ Solving for $v^2$, we have: $$v^2 = \frac{2 \times 8 \times 10^{-4}}{0.01} = 0.16$$ Thus, $v = \sqrt{0.16} = 0.4 \, \text{m/s}$.
The particle undergoes tangential acceleration, so we apply the equation for final velocity under constant acceleration: $$v = u + a_t t$$ where $u = 0$ (initial velocity), $v = 0.4 \, \text{m/s}$. The number of revolutions is 2, so the distance traveled is: $$s = 2 \times 2\pi R = 4\pi \times 0.064 \, \text{m} = 0.256\pi \, \text{m}$$
Using the equation for motion: $$s = ut + \frac{1}{2} a_t t^2$$ Simplifying, $$0.256\pi = \frac{1}{2} a_t t^2$$ As $v = a_t t$, $$0.4 = a_t \cdot t \implies t = \frac{0.4}{a_t}$$ Substituting t: $$0.256\pi = \frac{1}{2} a_t \left(\frac{0.4}{a_t}\right)^2 = \frac{0.08}{2a_t}$$ Solve for $a_t$: $$0.256\pi = \frac{0.08}{2a_t}$$ $$a_t = \frac{0.08}{2 \times 0.256\pi}$$ = 0.1 \text{m/s}^2

The correct magnitude of the tangential acceleration a_t is thus $0.1 \, \text{m/s}^2$.

Hence, the correct answer is $0.1 \, \text{m/s}^2$.

The Correct Option is D

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