Physics M.C.Q [1M]

1. $\mu<\mu',\text{M < M}'$

Explanation:

Let T be the force applied by the boy on the block.

Free body diagram for the box:

The condition for preventing the slide is

$\text{f}_{\text{max}}>\text{T}$

$\mu'\text{M}'\text{g}>\text{T} \ ...(\text{i})$

Now see the free body diagram of a boy of mass M:

$\text{f}_{\text{max}}=\mu\text{mg}$

The condition for preventing the slide is

$\text{f}_{\text{max}}>\text{T}$

$\mu\text{mg}>\text{T}$

The condition for sliding the entire system (block and boy) is f' > f (block is not slide)

$\mu'\text{M}'\text{g}>\mu\text{mg}$

$\mu'\text{M}'>\mu\text{m}$

$\mu<\mu'$

$\text{m < M}'$

1. F > F_N

2. F > f

4. F_N - f < F < F_N + f

Explanation:

The system is in equilibrium condition when F = f.

Hence, the net horizontal force is zero.

$\\\text{f}=\mu\text{F}_{\text{N}}$

$\text{F}>\text{F}_{\text{N}}$

$\text{f}=\text{F}_{\text{N}}$ and $0\leq\mu\leq1$

Therefore, we can say that F > f. So the net horizontal force is nonzero.

F > f, and so the net horizontal force is zero.

$\text{F}_{\text{N}}>\text{f}\Rightarrow\text{F}_{\text{N}}>\mu\text{F}_{\text{N}}\Rightarrow\mu<1$

Here, the given relation between F and f i.e.

$\text{F}>\text{f}$ and $\text{f}=\mu\text{F}_{\text{N}}$ will not be satisfied So it cannot be said that the net horizontal force is zero or nonzero.

$\text{F}_{\text{N}}-\text{f}<\text{F}<\text{F}_{\text{N}}+\text{f}$

$\because\text{f}=\mu\text{F}_{\text{N}}$

$\frac{\text{f}}{\mu}-\text{}f<\text{F}<\frac{\text{f}}{\mu}+\text{f}$

$\text{f}\Big(\frac{1-\mu}{\mu}\Big)<\text{F}<\text{f}\Big(\frac{1+\mu}{\mu}\Big)$

For the above relation, we can say that F ≠ f and so the net horizontal force is nonzero.

2. Smaller friction.

Explanation:

According to the first law of the limiting friction,

$\text{f}=\mu\text{N}$

where f is the frictional force

$\mu$ is the coefficient of friction

N is the normal reaction force

When we take smaller steps on ice, the normal reaction force exerted by the ice is small. Therefore, the smaller steps ensure smaller friction.

3.  Is same for both cars.

Explanation:

Given: both the cars have same initial speed.

Let the masses of the two cars be m₁ and m_2.

Frictional force on car with mass $\text{m}_1=\mu\text{m}_1\text{g}$ 

So, the deceleration due to frictional force $=\frac{\mu\text{m}_1\text{g}}{\text{m}_1}=\mu\text{g}$

Frictional force on car with mass $\text{m}_2=\mu\text{m}_2\text{g}$

So, the deceleration due to frictional force $=\frac{\mu\text{m}_2\text{g}}{\text{m}_2}=\mu\text{g}$

As both the acceleration are same, from the second equation of motion $\text{s = ut}+\frac{1}{2}\text{at}^2$

Thus, we can say that both the cars have same minimum stopping distance.

2. Decrease.

Explanation:

According to the first law of limiting friction,

$\text{f}=\mu\text{N}$

where f is the frictional force

N is the normal reaction force

$\mu$ is the coefficient of static friction

and

$\text{N = mg}-\text{F}\cos\theta$

where m is the mass of the body

F is the contact force acting on the body

If we decrease the angle between this contact force and the vertical, then $\text{F}\cos\theta$ increases and the normal reaction force (N) as well as the frictional force (f) decrease.

4. All of three are possible.

Explanation:

We know that

$\text{N = mg}\cos\theta^{\circ}$

$\text{f}_{\text{max}}=\mu\text{N}=\mu\text{mg}\cos\theta$

where N = normal reaction force

f_max = frictional force

$\theta$ = angle of inclination

$\mu$ = coefficient of friction

When the block just begins to slide, it means

$\text{mg}\sin\theta=\text{f}_{\text{max}}$

$\text{mg}\sin\theta=\mu\text{mg}\cos\theta$

$\mu=\tan\theta$

and the coefficient of friction depends on the angle of inclination $(\theta)$ and does not depend on mass.

Now consider the block sliding condition:

$\text{mg}\sin\theta-\text{f}_{\max}=\text{ma}$

$\text{mg}\sin\theta-\mu\text{mg}\cos\theta=\text{ma}$

$\therefore\text{a = g}(\sin\theta-\mu\cos\theta)$

From the above equation it is clear that acceleration does not depend on the mass but depends on $\theta$ and $\mu.$

3. $\text{Mg}\leq\text{F}\leq\text{Mg}\sqrt{1+\mu^2}$

Explanation:

Let T be the force applied on an object of mass M.

If T = 0, F_min = Mg.

If T is acting in the horizontal direction, then the body is not moving.

$\therefore\text{T}=\mu(\text{mg})$

$\text{F}_{\text{max}}=\sqrt{(\text{Mg})^2+(\text{T})^2}$

$=\sqrt{(\text{Mg})^2+(\mu\text{Mg})^2}$

Thus, we have:

$\text{Mg}\leq\text{F}\leq\text{Mg}\sqrt{1+(\mu)^2}$

4. > 500N for time $\triangle\text{t}_1$ and $\triangle\text{t}_3$ and 500N for $\triangle\text{t}_2.$

Explanation:

During the time interval $\triangle\text{t}_2,$ the scooter is moving with a constant velocity, which implies that the force exerted by the seat on the man is 500N (for balancing the weight of the man).

During the time interval $\triangle\text{t}_1$ and $\triangle\text{t}_3,$ the scooter is moving with constant acceleration and deceleration, which implies that a frictional force is also applied. Therefore, the net force exerted by the seat on the man should be > 500N.