Test: Free Body Diagrams (June 4) - JEE MCQ

We know that F = ma, and for m = 1kg and a = 1m/s²
We get F = 1N which is one newton

At first considering both blocks as one system with only one external force F
We get common acceleration at right be a = 100/15 m/s²
Now considering 10 kg block
We get F - T = 10a
i.e. T = 100  - 10(100/15)
= 100 (1 - 2/3)
= 33.33 N

When an object falls freely, it experiences weightlessness. This is because the object and the load on it are both accelerating towards the ground at the same rate due to gravity. Therefore, the object and the load on it will have the same weight as they would have if they were stationary on the ground.

In this case, the man is carrying a load of 10 kgf on his head and jumps from a tower. As he falls freely, both the man and the load on his head will experience weightlessness. Therefore, the weight of the load experienced by the man will be zero.
 

For a body able to loop a vertical circle, the tension in the string must not be lesser than zero at the highest point. Hence when the tension at the highest point is zero we will obtain the minimum velocity. I.e.
At the highest point as T = 0, hence we get
Mg = Mv²/R
(comparing the centripetal acceleration with weight, where R is radius of vertical circle)
Hence we get the minimum velocity at the topmost point to be 
Thus if we apply the work energy theorem upon the particle from its lowermost point to its top most point, we get net displacement as 2R and thus minimum velocity at lowermost point to be 

Weight of the block, mg = 5kg x 10 m/s² = 50 N.
According to Newton’s third law, the action of the block, that is the force exerted on the floor by the block is equal to 50 N in magnitude and is directly vertically downward.

As the lift itself is accelerating hence the net downwards acceleration of the ball will be 9.8 - 3
= 6.8 m/s²
Thus by applying second equation of motion we get, t = 
 = 0.71

As the lift moves upwards but the spring feels itself at rest hence we need to compensate the non inertial frame by adding an appropriate pseudo force to treat it as an inertial frame. Hence the pseudo force to be applied acts on every mass in the lift which is equal to mass x acceleration (=g) downwards.
Hence the tension in the spring would be 40N (20 due to weight and 20 pseudo). Thus the reading would be 4kg.

When the force F is applied at some angle, only the component of it which is parallel to the surface would provide acceleration to it. The other one would be balanced by the normal force. Hence as the component parallel to the surface is F.cosΘ , we get a = F.cosΘ / m

At first consider both blocks as one system then we get the common acceleration of the system,  a =  F / m₁ + m₂
Now if we see only mass m1
We get that for contact force f,
F - f = m₁a
Thus we get f = F - m₁ F / m₁ + m₂
=  m₂F /  m₁ + m₂

Both of them are vector quantities. And both of them can be easily simplified. If taken in the vector form then the task is even easier. Thus it is not necessary for the force or the couple to be vector only, even if the magnitude is taken, the simplification is done in the 2D.