Question

A block weighing $10N$ travels down a smooth curved track $AB$ joined to a rough horizontal surface (figure). The rough surface has a friction coefficient of $0.20$ with the block. If the block starts slipping on the track from a point $\text{1}\text{.0 m}$ above the horizontal surface, the distance it will move on the rough surface is:

A) $5.0m$

B) $10.0 m$

C) $15.0 m$

D) $20.0m$

Answer 1

Hint: From path A to B, since only gravitational force is acting on the body. So the work done on the body is independent of the path of the body's motion up to point B. Then, from point B, since the path is rough, friction comes into play during the block’s motion through it. Thus, for path AB, energy conservation principle can be applied and after point B, for the rest path of motion, a work-energy theorem should be used to solve the problem.

Formula used:

Energy conservation principle: When only conservative forces are acting on a body, then this theory can be applied on the body. This theory states:

$K.E{{.}_{i}}+P.E{{.}_{i}}=K.E{{.}_{f}}+P.E{{.}_{f}}$

Where, $K.E{{.}_{i}}$ is the initial kinetic energy of the body, $P.E{{.}_{i}}$ is the initial potential energy of the body, $K.E{{.}_{f}}$ is the final kinetic energy of the body and $P.E{{.}_{f}}$ is the final potential energy of the body.

Work energy principle: When non-conservative forces, like friction, act on a body, then this theory should be used. This theory states:

Work done by all the forces on a body,

$\Delta E={{E}_{f}}-{{E}_{i}}$

Where, ${{E}_{f}}$ is the final total energy of the body and ${{E}_{i}}$ is the initial total energy of the body.

Complete step-by-step answer:

For motion of block through path $AB$, applying energy conservation principle at points $A$ (initial state) and $B$ (final state) for the body, taking datum at level of $B$.

${{\left( K.E. \right)}_{A}}+{{\left( P.E. \right)}_{A}}={{\left( K.E. \right)}_{B}}+{{\left( P.E. \right)}_{B}} $

$\dfrac{1}{2}mv_{A}^{2}+mg{{h}_{A}}=\dfrac{1}{2}mv_{B}^{2}+mg{{h}_{B}} $

Since ${{v}_{A}}=0,{{h}_{A}}=1.0m,{{v}_{B}}=v,{{h}_{B}}=0m $

So, we get

$0 + 10 \times 1.0 = \dfrac{1}{2} \times \dfrac{10}{10} \times v^2 + 0 $

$\Rightarrow v=\sqrt{20}\text{ m}{{\text{s}}^{-1}}$

For motion of the body after point B, let the body travel distances $s$ before stopping, applying work-energy principle on the body after point B (initial state) and just before coming to rest (final state).

${{W}_{f}}+{{W}_{g}}=T.E{{.}_{f}}-T.E{{.}_{i}}={{\left( K.E.+P.E. \right)}_{f}}-{{\left( K.E.+P.E. \right)}_{i}}$

Where, ${{W}_{f}}$ is the work done by friction force on the body $\left( {{W}_{f}}=-\mu mgs \right)$ and ${{W}_{g}}$ is the work done by gravitational force on the body $\left( {{W}_{g}}=0 \right)$.

So, we get

$-\mu mgs+0=0-\dfrac{1}{2}mv_{B}^{2}\text{ }\left( \because T.E{{.}_{f}}=0,T.E{{.}_{i}}=\dfrac{1}{2}mv_{B}^{2} \right) $

$\text{ }{{v}_{B}}=\sqrt{20}\text{ m}{{\text{s}}^{-1}}\text{, }\mu =0.20 $

$\Rightarrow s=\dfrac{v_{B}^{2}}{2\mu g}=\dfrac{20}{2\times 0.20\times 10}=5\text{ m}\text{.}$

Thus, the block will move a distance of $5\text{ m}$ on the rough surface before coming to rest.

Hence, the correct answer is $\left( A \right)$.

Note: Students should be careful while attempting these types of problems as they may think it of a kinematics problem and thus get distracted in a wrong way. Problems involving friction force and distance travelled, are easy to be assessed using the work-energy principle.