Laws of Motion - 2 - Free MCQ Practice Test with solutions, JEE Physics

Here, m₁​a=T−m_1​g...(1)
m_2​a=m₂​g−T...(2)
(1)+(2),a=m_1​+m₂​m₂​−m₁​​g
Acceleration of m₁​, a₁​=−a= ​m_1​−m₂​​g/ m₁​+m₂
For this calculation, we assume that the string is massless and rim of the pulley is frictionless otherwise there would have been a frictional force and mass of string included in the acceleration expression. The tension will come due to the inextensible string, otherwise. the elastic effect would have to be included in the tension. The mass of the pulley does not affect the given expression as tension on both the sides remain the same. Thus, option C is correct.
 

Let the mass of object launched is m, the acceleration due to gravity is g and the acceleration of the object at any instant = a
Using Newton’s second law of motion,




So,



Only option B follows the above discussion

Here m₂​>m₁​
If lighter man is stationary,
Vertical force on m_1​
T−m_1​g=0
 ⇒T=m₁​g
Vertical force on m₂​
m₂a=T−m₂​g=m₁​g−m_2​g
a= ​m₁/ m₂ ​​g−g
since  ​m₁/ m₂​​<1, so in this case, a is negative. thus the heavier man of mass m₂​ is sliding down.
If heavier man is stationary,
Vertical force on m₂​=T−m₂​g=0
⇒T=m₂​g
Vertical force on m_1​=m_1​a=T−m_1​g=m₂​g−m₁​g
a= ​m₂/ m₁ ​​g−g
since m_2​>m₁​ so in this case, a is positive . Thus the lighter man will climb up with some acceleration.
From the above two cases we can say that they will move with the same acceleration over the pulley but in the opposite direction.
Ans:(A),(B),(D)
 

(C) 2 only

[2] The weight of the 5kg mass acts vertically downwards

For an inertial frame the acceleration of the object should be = 0

a=F/m=αt/m……(i)
or a ∝ t
i.e. a-t graph is a straight line passing through
origin.
If u=0, then integration of Eq. (i) gives,
v=αt²/2m or v∝t²
Hence in this situation (when u=0) v-t graph is a
parabola passing through origin.

Force - force of friction = mass × acceleration
250 - f = 30 × 4
f = 250 – 120
= 130N

For force to be zero net acceleration should be zero i.e. velocity should remain constant. In an x-t graph if slope is a straight line it means that velocity is constant which is the case for regions AB, BC and CD.

It not a necessity that F₂ must be or must not be equal to F₁. If F₂ is opposite in direction but smaller/larger in magnitude, it will still stop the particle.

If we move relative to the block, then

Now, fr = μmg
a = μg
a = - 2 m/s2 (i.e. towards right)
The final velocity of the small block is 0 so
2as = vf2 – vi2 
-4s = -16
s = 4m

The weight of the mass of block, m = 2.5 kg
Now comparing the values of the weight of block = size of Normal force = mg = (2.5)(10) = 25 N
The horizontal force applied = 15 N (capable to move blocks)
{where d = 10 m, and t = 5 s, and "a" = block's acceleration}
The net force acting on the block of weight = ma = (2.5)(0.8) = 2.0 N
The force required to oppose the external force = 15 - 2 = 13 N
The value of the Kinetic Friction Force = 13 N
Therefore, dividing the Kinetic Force Friction and Horizontal Force by Normal Force we get
Coefficient of kinetic friction = 13/25 = 0.52 ANS
Coefficient of static friction = 15/25 = 0.60 ANS
 

f_1max = 0.2 × 40 = 8 N
Maximum acceleration of 4 kg block
= f_1 max / 4kg
= 2 m/s²

The greater the force exerted on the ground, the greater is the reaction and therefore the greater pull will be given to the rope.

Let the mass of the man be m and that of the block be M, where M > m.
The coefficient of friction between both (man and block) and the ground is μ.

The maximum frictional force on the man = μm_g
The maximum frictional force on the block = μM_g

Since M > m, the block experiences a greater maximum frictional force.
Hence, the man cannot apply a pull on the block greater than μm_g without himself starting to move, while the block will start moving only if the applied tension exceeds μM_g.
Because μM_g > μm_g, the man alone cannot make the block move while he remains stationary.
Therefore, the block will not move unless the man also moves.
→ Assertion A is correct.

When both start moving, the same tension T acts on the man and the block.
For the block: T− μM_g = Ma₂
For the man: μm_g − T = ma₁

Since the same tension acts on a smaller mass (the man),

a₁ > a₂

Hence, if both move, the acceleration of the man is greater than that of the block.
→ Assertion C is correct.

The friction force on the mass M due to rough surface is f = μMg
In equilibrium, 
For mass m,
T = mg,
for mass M,
f = μMg = T
⇒ mg = μMg
⇒ m = μM
If mg > μMg, then there would be an acceleration.
Thus, we get m > μM

That is the statement of Newton's 2nd law of motion.

If no force is applied, the block A will slip on C towards right and the block B will move downward. Suppose the minimum force needed to prevent slipping is F, taking A+B+C as the system, the only external horizontal force on the system is F. Hence the acceleration of the system is
a = F/2m+M
take block A as the system. The force on A are
 1) Tension T by the string towards right
 2) Friction f by the block C towards left
 3) Weight mg downward and
 4) Normal force N upward.
For vertical equilibrium N = mg.
As the minimum force needed to prevent slipping is applied, the friction is limiting thus
f = μN = μmg
As the block moves towards right with an acceleration a,
T –f = ma
Or T - μmg = ma                                                               ----------------- (2)
Now take the block B as the system the forces are as shown in fig (b).      
 1) Tension T upward.
 2) Weight mg downward
 3) Normal force N1 towards right
 4) Friction f1 upward.
As the block moves towards right with an acceleration a, N = ma.
As the friction is limiting, f1= μN’ = μma
For vertical equilibrium
T + f1 = mg
T + μma = mg 
Eliminating T from (1) and (3)
amin = (1- μ)*g/(1+ μ)
When a large force is applied the block A slips on C-towards left and the block B slips on C in the upward direction. The friction on A is towards right and that on B is downwards solving as above, the acceleration in this case is
Thus, the force should be between
(1- μ)(M+2m)*g/(1+ μ) and
(1+ μ)(M+2m)*g/(1- μ)
 

The diagram should be changed to the following:

The question is incomplete and is too vague to be found
It should be removed so as to not cause confusion.

Since the force is uniform the particle takes equal time to got to E from P and to Q from E. Now if T is the time taken by it to go to E from P then the particle takes 4 such time periods to come back to its original position.

F=ma ⇒ a=F/m
If time taken by the particle to move from p to E is t then,
A=1/2 aT2=1/2(F/m)T²
⇒T= √2am/F
Time taken by the particle to come back to P is t=4T
⇒ t=4(√2Am/F)

In the velocity time graph, the slope value should give the value of acceleration.