Physics M.C.Q (1 Marks)

2. $2\text{E}$

Explanation:

Let x_A and x_B be the extensions produced in springs A and B, respectively.

Restoring force on spring A, F = k_Ax_{A ...(1)}

Restoring force on spring B, F = k_Bx_B ...(2)

From (1) and (2), we get:

k_Ax_A = k_Bx_B

It is given that k_A = 2k_B

$\therefore\ \text{x}_\text{B}=2\text{x}_\text{A}$

Energy stored in spring A:

$\text{E}=\frac{1}{2}\text{k}_\text{A}\text{x}_\text{A}^2\ \dots(3)$

Energy stored in spring B:

$\text{E}'=\frac{1}{2}\text{k}_\text{B}\text{x}_\text{B}^2=\frac{1}{2}\Big(\frac{\text{k}_\text{A}}{2}\Big)(2\text{x}_\text{A})^2$

$\therefore\ \text{E}'=2\times\Big(\frac{1}{2}\text{k}_\text{A}\text{x}_\text{A}^2\Big)=2\text{E}$ [From (3)]

3. $\text{mgvt}\sin^2\theta$

Explanation:

Distance (d) travelled by the elevator in time t = vt

The block is not sliding on the wedge.

Then friction force $(\text{f})=\text{mg}\sin\theta$

Work done by the friction force on the block in time t is given by,

$\text{W}=\text{Fd}\cos(90-\theta)$

$\Rightarrow\text{W}=\text{mg}\sin\theta\times\text{d}\times\cos(90-\theta)$

$\Rightarrow\text{W}=\text{mgd}\sin^2\theta$

$\therefore\ \text{W}=\text{mg}\nu\text{t}\sin^2\theta$

4. $-\frac{1}{4}\text{kx}^2$

Explanation:

The work done by the spring on both the masses is equal to the negative of the increase in the elastic potential energy of the spring.

The elastic potential energy of the spring is given by $\text{E}_\text{p}=\frac{1}{2}\text{kx}^2.$

Work done by the spring on both the masses $=-\frac{1}{2}\text{kx}^2$

$\therefore$ Work done by the spring on each mass $=\frac{1}{2}\Big(-\frac{1}{2}\text{kx}^2\Big)$

$=-\frac{1}{4}\text{kx}^2$

For an elastic spring, the work done is equal to the negative of the change in its potential energy.

When the spring was initially compressed or stretched by a distance x, its potential energy is given by,

$(\text{P.E.})_\text{i}=\frac{1}{2}\text{kx}^2$

When it finally comes to its natural length, its potential energy is given by,

$(\text{P.E.})_\text{f}=0$

$\therefore$ Work done $=-[(\text{P.E.})_\text{f}-(\text{P.E.})_\text{i}]=-\Big[0-\frac{1}{2}\text{kx}^2\Big]$

$=\frac{1}{2}\text{kx}^2$

1. Must depend on the speed of projection.
2. Must be larger than the speed of projectio.

Explanation:

Consider that the stone is projected with initial speed v.

As the stone is falls under the gravitational force, which is a conservative force, the total energy of the stone remains the same at every point during its motion.

From the conservation of energy, we have:

Initial energy of the stone = final energy of the stone

i.e., (K.E.)_i + (P.E.)_i = (K.E.)_f + (P.E.)_f

_{$=\frac{1}{2}\text{mv}_\text{r}^2+\text{mgh}=\frac{1}{2}\text{m}(\text{v}_\text{max})^2$}

_{$\Rightarrow\text{v}_\text{max}=\sqrt{\text{v}^2+2\text{gh}}$}

From the above expression, we can say that the maximum speed with which the stone hits the ground depends on the speed of projection and greater than it.

3. $\sqrt{3\text{gl}}$

Explanation:

Suppose that one end of an extensible string is attached to a mass m, while the other end is fixed. The mass moves with a velocity v in a vertical circle of radius R. At some instant, the string makes an angle $\theta$ with the vertical as shown in the figure.

For a complete circle, the minimum velocity at L must be $\nu_\text{L}=\sqrt{5\text{gl}}.$

Applying the law of conservation of energy, we have:

Total energy at M = total energy at L

i.e., $\frac{1}{2}\text{m}\nu_{\text{M}^2}+\text{mgl}=\frac{1}{2}\text{m}\nu_{\text{L}^2}$

$\Rightarrow\frac{1}{2}\text{m}\nu_{\text{M}^2}=\frac{1}{2}\text{m}\nu_{\text{L}^2}-\text{mgl}$

Using $\nu_\text{L}\geq\sqrt{5\text{gl}},$ we have:

$\frac{1}{2}\text{m}\nu_{\text{M}^2}\geq\frac{1}{2}\text{m}(5\text{gl})-\text{mgl}$

$\therefore\ \nu_\text{M}=\sqrt{3\text{gl}}$

4. The speed does not depend on the initial direction.

Explanation:

As the stone falls under the gravitational force, which is a conservative force, the total energy of the stone remains the same at every point during its motion.

From the conservation of energy, we have:

Initial energy of the stone = final energy of the stone

i.e., (K.E.)_i+ (P.E.)_i = (K.E.)_f+ (P.E.)_f

$=\frac{1}{2}\text{mv}^2+\text{mgh}=\frac{1}{2}\text{m}(\text{v}_\text{max})^2$

$\Rightarrow\text{v}_\text{max}=\sqrt{\text{v}^2+2\text{gh}}$

From the above expression, we can say that the maximum speed With which stone hits the ground does not depend on the initial direction.

3. Potential energy.

Explanation:

The negative of the work done by the conservative internal forces on a system is equal to the changes in potential energy.

i.e. $\text{W}=-\triangle\text{ P.E.}$