Question

A body is projected up along an inclined plane from the bottom with speed ${\mathrm{v}}_1$. if it reaches the bottom of the plane with a velocity of ${\mathrm{v}}_2$, find $\frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}$ if $\mathrm{\theta }$ is the angle of inclination with the horizontal and $\mathrm{\mu }$ be the coefficient of friction.

• A. $\frac{\mathrm{sin\theta }+\mathrm{\mu cos\theta }}{\mathrm{sin\theta }-\mathrm{\mu cos\theta }}$
• B. $\sqrt{\frac{\mathrm{sin\theta }+\mathrm{\mu cos\theta }}{\mathrm{sin\theta }-\mathrm{\mu cos\theta }}}$
• C. $\frac{\mathrm{cos\theta }+\mathrm{\mu sin\theta }}{\mathrm{cos\theta }-\mathrm{\mu sin\theta }}$
• D. $\sqrt{\frac{\mathrm{cos\theta }+\mathrm{\mu sin\theta }}{\mathrm{cos\theta }-\mathrm{\mu sin\theta }}}$

Answer 1

The correct option is B

$\sqrt{\frac{\mathrm{sin\theta }+\mathrm{\mu cos\theta }}{\mathrm{sin\theta }-\mathrm{\mu cos\theta }}}$

Step 1. Given data.

We have given the speed ${\mathrm{v}}_1$, and velocity ${\mathrm{v}}_2$ with angle $\mathrm{\theta }$ and coefficient friction is $\mathrm{\mu }$.

We have to find the ratio of $\frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}$.

Step 2. Concept used.

The law of conservation of energy states that the total energy of an isolated system is constant. Energy is neither created nor destroyed, it can only be transformed from one to another or transferred from one system to another.

The equation of motion is,

${\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{as}$

Here, $\mathrm{v}$ is final velocity, $\mathrm{u}$ is initial velocity, $\mathrm{a}$ is acceleration and $\mathrm{s}$ is displacement.

Step 3. Find equation for speed

From the above figure, there is the distance travelled by the body along the inclined plane is the same.

Use the law of conservation of energy for upward motion.

We know that, concept of net force and its relationship to mass and acceleration.

${\mathrm{a}}_{\mathrm{eff}}=\frac{{\mathrm{F}}_{\mathrm{net}}}{\mathrm{m}}$

Here, ${\mathrm{F}}_{\mathrm{net}}$ is net force, $\mathrm{m}$ is mass and $\mathrm{a}$ is acceleration.

So,

${\mathrm{a}}_{\mathrm{eff}}=\frac{{\mathrm{F}}_{\mathrm{net}}}{\mathrm{m}}$

$=\frac{-\mathrm{mg}\left(\mathrm{sin\theta }+\mathrm{\mu cos\theta }\right)}{\mathrm{m}}$

$=-\mathrm{g}\left(\mathrm{sin\theta }+\mathrm{\mu cos\theta }\right)$

Use equation of motion, we get,

${\mathrm{v}}^2-{{\mathrm{v}}_1}^2=2{\mathrm{a}}_{\mathrm{eff}}\mathrm{s}$

$0^2-{{\mathrm{v}}_1}^2=2{\mathrm{a}}_{\mathrm{eff}}\mathrm{s}$

${{\mathrm{v}}_1}^2=-2\left(\mathrm{g}\left(\mathrm{sin\theta }+\mathrm{\mu cos\theta }\right)\right)\mathrm{s}$ ------- $\left(1\right)$

Step 4. Find the equation for velocity

Now, for down hill,

$\mathrm{a}{\text{'}}_{\mathrm{eff}}=\frac{{\mathrm{F}}_{\mathrm{net}}}{\mathrm{m}}$

$=\frac{\mathrm{mg}\left(\mathrm{sin\theta }-\mathrm{\mu cos\theta }\right)}{\mathrm{m}}$

$=\mathrm{g}\left(\mathrm{sin\theta }-\mathrm{\mu cos\theta }\right)$

Use equation of motion, we get,

${\mathrm{v}}^2-{{\mathrm{v}}_2}^2=2\mathrm{a}{\text{'}}_{\mathrm{eff}}\mathrm{s}$

$0^2-{{\mathrm{v}}_2}^2=2\mathrm{a}{\text{'}}_{\mathrm{eff}}\mathrm{s}$

${{\mathrm{v}}_2}^2=2\left(\mathrm{g}\left(\mathrm{sin\theta }-\mathrm{\mu cos\theta }\right)\right)\mathrm{s}$ ------- $\left(2\right)$

Step 5. Find the ratio

Divide equation $\left(1\right)$ by $\left(2\right)$, we get,

${\left(\frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}\right)}^2=\frac{2\mathrm{g}\left(\mathrm{sin\theta }+\mathrm{\mu cos\theta }\right)}{2\mathrm{g}\left(\mathrm{sin\theta }-\mathrm{\mu cos\theta }\right)}$

$\frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}=\sqrt{\frac{\left(\mathrm{sin\theta }+\mathrm{\mu cos\theta }\right)}{\left(\mathrm{sin\theta }-\mathrm{\mu cos\theta }\right)}}$

Hence, option $\mathrm{B}$ is the correct answer.