Newton S Laws With Friction And Beyond(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

NEWTON S LAWS WITH FRICTION AND BEYOND MCQs

Total Questions : 15 | Page 1 of 2 pages

A civil engineer wishes to redesign the curved roadway in such a way that a car will not have to rely on friction to round the curve without skidding. In other words, a car moving at the designated speed can negotiate the curve even when the road is covered with ice. Such a ramp is usually banked, which means that the roadway is tilted toward the inside of the curve. Suppose the designated speed for the ramp is to be 10.0 m/s and the radius of the curve is 20.0 m. At what angle should the curve be banked?

$30^{\circ}$
$t a n^{- 1} \frac{1}{3}$
$60^{\circ}$
$t a n^{- 1} \frac{1}{2}$

Discuss Question

Answer: Option D. ->

t a n^{- 1} \frac{1}{2}

:
D

A Civil Engineer Wishes To Redesign The Curved Roadway In Su...

\sum F_{r} = N s i n θ = \frac{m v^{2}}{r}

\sum F_{g} = N c o s θ - m g = 0

\Rightarrow t a n θ = \frac{v^{2}}{r g}

\Rightarrow θ = t a n^{- 1} (\frac{10^{2}}{(20) (10)}) = t a n^{- 1} (\frac{1}{2})

Question 2.

A scooter starting from rest moves with a constant acceleration for a time $Δ$ $t_{1}$ , then with a constant velocity for the next $Δ$ $t_{2}$ , and finally with a constant deceleration. A person of mass 50 kg on the scooter, behind the driver manages to stay at rest with respect to the scooter without touching any other part. The force exerted by the seat on the man is

Take g=10m/ $s^{2}$

500 N throughout the journey
less than 500 N throughout the journey
more than 500 N throughout the journey
> 500 N for time $△$ $t_{1}$ and $△$ $t_{3}$ , and 500 N for $△$ $t_{2}$ ,

Discuss Question

Answer: Option D. -> > 500 N for time

△

t_{1}

and

△

t_{3}

, and 500 N for

△

t_{2}

,
:
D

So when scooter is accelerating the man is also accelerating along with it so friction from the scooter's seat on man is making him move forward.

Also there is a Normal force

N = mg = 500 N

So Net force while accelerating will be $\sqrt{{(500)}^{2} + {(f_{r})}^{2}}$ > 500

Same is true about decelerating

So while accelerating or decelerating the force on man by scooter is more than 500N. When the scooter is going with constant velocity, the man is also going with constant velocity. So 0 acceleration in horizontal direction so no net force in horizontal direction so no friction.

Only Normal N = mg = 500

Question 3.

In a situation the contact force by a rough horizontal surface on a body placed on it has constant magnitude. If the angle between this force and the vertical is decreased, the frictional force between the surface and the body will

Increase
decrease
Remain the same
may increase or decrease

Discuss Question

Answer: Option B. -> decrease
:
B

Now if the angle made by $F_{c}$ and the vertical decreases then horizontal componenet decreases

As you can see the horizontal component decreases, Which in this case turns out to be frictional force.

Yes the magnitude of frictional force depends upon the angle made by Contact force and the Vertical/Horizontal

Question 4.

A body of mass M is kept on a rough horizontal surface (friction coefficient = $μ$ ). A person is trying to pull the body by applying a horizontal force but the body is not moving. The force by the surface on the body is F, where

F = Mg
F = $μ$ Mg
Mg $\leq$ F $\leq$ Mg $\sqrt{1 + μ^{2}}$
Mg $\geq$ F $\geq$ Mg $\sqrt{1 + μ^{2}}$

Discuss Question

Answer: Option C. -> Mg

\leq

\leq

\sqrt{1 + μ^{2}}

:
C

If $F_{1}$ = 0

So contact force will only have Normal which is equal to Mg
So F will be Mg

Now as $F_{1}$ increases fr keeps increasing till it reaches limiting value i.e., $μ$ N = $μ$ mg
So, in that case, contact force F will be

$\sqrt{N^{2} + {f_{r}}^{2}}$ = $\sqrt{{(M g)}^{2} + {(μ M g)}^{2}}$

F = Mg $\sqrt{1 + μ^{2}}$

This means contact force F will lie between Mg $\leq$ F $\leq$ Mg $\sqrt{1 + μ^{2}}$

Question 5.

A body of mass 400g slides on a rough horizontal surface. If the frictional force is 3.0 N, find (a) the angle made by the contact force on the body with the vertical and (b) the magnitude of the contact force. Take g = 10 $\frac{m}{s^{2}}$

$37^{\circ}$ , 5N
$37^{\circ}$ , 4N
$53^{\circ}$ , 4N
$53^{\circ}$ , 5N

Discuss Question

Answer: Option A. ->

37^{\circ}

, 5N
:
A

Let the contact force on the block by the surface be F which makes an angle $θ$ with the vertical

The component of F perpendicular to the contact surface is the normal reaction N and the component of F parallel to the surface is the well-known frictional force f. As the surface is horizontal, N is vertically upward. For vertical equilibrium,

N = mg = (0.400 kg) $\frac{10 m}{s^{2}}$ = 4.0 N.

The frictional force is f = 3.0 N.

(a)

Or, θ = ${t a n}^{- 1}$ ( $\frac{3}{4}$ ) = $37^{\circ}$ .

(b)The magnitude of the contact force is

F = $\sqrt{N^{2} + f^{2}}$

= $\sqrt{{(4.0 N)}^{2} + {(3.0 N)}^{2}}$ = 5.0 N.

Question 6.

Block A is kept on top of block B and moves leftwards with respect to B. Assume the friction to be present between the blocks. The direction of friction force on A by B, will be?

$\to$ (Rightwards)
$\leftarrow$ (Leftwards)
$↘$
$↙$

Discuss Question

Answer: Option A. ->

\to

(Rightwards)
:
A

Block A moves in the leftward direction with respect to B. ∴ Friction will act in the rightward direction on A. As the bonds formed will oppose the relative motion.

Question 7.

A body of mass 10 $k g$ is slipping on a rough horizontal plane and moves with a deceleration of 4.0 $m / s^{2}$ What is the magnitude of frictional force?

40 N
20 N
100 N
Data insufficient (require coefficient of kinetic friction)

Discuss Question

Answer: Option A. -> 40 N
:
A

$f_{r}$ = ma

$f_{r}$ = 10 x 4 kg-m/ $s^{2}$

$f_{r}$ = 40 N

As friction can be the only external force acting of the body in horizontal direction.

Question 8.

A block is placed on a rough floor and a horizontal external force F is applied on it. The force of friction f by the floor on the block is measured for different values of F and a graph is plotted between them. (taking f on Y and F on X axis)

The graph is a straight line of slope $45^{\circ}$
The graph is a straight line parallel to the F-axis
The graph is a straight line of slope $45^{\circ}$ for small F and a straight line parallel to the F-axis for large F
There is a small kink on the graph

Discuss Question

Answer: Option C. -> The graph is a straight line of slope

45^{\circ}

for small F and a straight line parallel to the F-axis for large F
:
C and D

Limiting friction = $μ_{s}$ mg

So till F $\leq$ $f_{r}$

F = fr

So straight line at angle $45^{\circ}$

After that even if Force F increases the value of friction will be fixed at $μ_{K}$ mg

So straight line also $μ_{K}$ < $μ_{s}$

So there will be kink.

Question 9.

Spring fitted doors close by themselves when released. You want to keep the door open for a long time, say for an hour. If you put a half kg stone in front of the open door, it does not help. The stone slides with the door and the door gets closed. However, if you sandwich a 20 gm piece of wood in the small gap between the door and the floor, the door stays open. Explain why a much lighter piece of wood is able to keep the door open while the heavy stone fails. Assume coefficient of friction between stone and ground and wood and ground are the same.

There is no horizontal force on the wood by the door
Surface area of contact between wood and door is less
Limiting friction will increase as the horizontal force on the block by the ground will increase
None of these

Discuss Question

Answer: Option C. -> Limiting friction will increase as the horizontal force on the block by the ground will increase
:
C

In this case, the door will apply a downward force on the wooden piece which will increase the normal force that the ground applies on the wooden piece. ∴ Limiting value of friction will increase.

Question 10.

In the following figure force F is gradually increased from zero. Draw the graph between applied force F and tension T in the string. The coefficient of static friction between the block and the ground is $μ_{s}$ .

Discuss Question

Answer: Option B. ->

:
B

As the external force F is gradually increased from zero it is compensated by the friction and the string bears no tension. When limiting friction is achieved by increasing force F to a value till $μ_{s}$ mg, the further increase in F is transferred to the string.