Test: Free Body Diagram- 2 - Grade 12 MCQ

Force on A’ and B’ is 40 N and 60 N respectively. Net force of the situation is 40 N upward. So to balance the force on A’ a force of 40 N has to apply on C’. Similarly a force of 100 N is also be applied to get the net force of 40 N in upward direction.

The initial velocity is 5 m/s and the block stops after ½ sec,
Hence the net negative acceleration is 10 m/s2
The net force active downwards parallel the plane is mg sin 30° + kmg cos 30°
Hence we get a = g ( sin 30° + kcos 30°) = 10
Thus we get  sin 30° + kcos 30° = 1
k = sec 30° - tan 30°
= 2/1.73 - 1/1.73
= 1/1.73
= 0.6

In the given condition, we get that a component of applied force acts parallel to force rightwards and one is perpendicular. Also the weight, normal and friction (leftwards) would also act upon it
Hence the correct Ans is D.

For a body to move with constant velocity the net force on body should be zero.Only option A can have a net force zero rather b,c net horizontal force is non-zero and in option d there is no horizontal force so no motion in horizontal direction So option a is most appropriate

The mass of the lift is 1000kg, the acceleration is 1m/s² and the acceleration due to gravity is 9.8m/s².

The tension developed in string is given as,

T = mg+ma

= 1000×9.8+1000×1

= 10800N

Thus, the tension developed in the string is 10800N.

Efficiency = output/input = (75×3) / (25×12)

=75/100 =75%

Let F be the min. force required to move the cart
then F = μN (along x axis )
Δ N = mg (along y x axis )
⇒ F = μmg = 1/3 x 60 x 9.8
⇒ F = 196 N.

The force parallel to the plane but downwards, W = mg.sin 60 = 980 x √3/2 = 848.7N
Maximum friction force acting, f = mg.cos 60 x 0.5 = 980 x ½ x 0.5 = 245N
Thus the minimum extra force required, let say F = 848.7 - 245 = 603.3 N

Considering the three block system, we have
100 = 60.a 
where a is the common acceleration of all the blocks
a = 5/3 m/s²
Considering mass 1 we get,
T₁ = 10 a
T₁ = 50/3 N
Considering mass 1 and 2 we get,
T₂  = 30 a
T₂ = 50N

On the block A, the tension acts leftwards while friction acts rightwards due to normal upwards which balances weight downwards. Similarly the tension acts upwards on block B which balances its weight.