Test: Solving Problems in Mechanics (NCERT) - NEET MCQ

Force will be same for all the surfaces.

Here,
Mass of the man, M = 55 kg
As the man is standing stationary w.r.t.  the belt,
Acceleration of man = Acceleration of belt
Acceleration of man a = 1ms⁻² 
Net force on the man,
F = Ma = (55 kg) × (1 m s^-2) = 55N
Force on man is 55N 

Here,
Mass of hehcopter, M = 2000 kg
Mass of the crew and passengers, m = 500 kg
Vertically upwards acceleration, a = 15 ms^-2
g = 10 m s^-2
(i) Force on the floor by the crew and passengers
= m (g + a) = 500 kg (10 + 15) m s^-2
= 12500 N = 1.25 x 10⁴ N
It acts vertically downwards.
(ii) Action of the rotor of the hehcopter on the surrounding air = (M + m) (g + a)
= (2000 + 500) kg (10 + 15) m s ^-2
= 62500 N = 6.25 x 10⁴ N
It acts vertically downwards
(iii) Force on the helicopter due to the surrounding air is equal and opposite to the action of the rotor of the helicopter on the surrounding air.
Force on the helicopter due to the surrounding air
= 6.25 x 10⁴ N
It acts vertically upwards.

Here, v = 20m s⁻¹, a = 2m s⁻², g = 10m s⁻²
The coin will fall back into the person’s hand after t s.
∴  t = 
= 10/3s

The reading on the scale is a measure of the force on the floor by the person. By the Newton’s third law this is equal and opposite to the normal force N on the person by the floor.
∴ When the lift is descending downward with a acceleration of a 9 ms⁻², then
N - 50 × 10 = 50 × 9
or N = 50 × 10 + 50 × 9 = 50(10 + 9)
= 950 N
∴  The reading of weighing machine is 95kg.

As the system is in equilibrium,
∴ T = W …(i)
and T = mg sin30^∘ …(ii)
From (i) and (ii), we get

W = mg sin 30^∘ = (100 N) × 1/2 = 50 N

Here, m(1) = 10kg, m = 20kg, F = 600N
Let T be tension of the string and a be common acceleration of the system.
a = 
= 20 ms^-2
When force F is applied on 10kg block, then the tension in the string is
T₂ = m₂a = (20kg)(20ms−²) = 400N

Given

Mass of body (m₂) = 10kg

Mass of body (m₁) = 20kg

Force applied =600N

Let a is the acceleration of the system.

When force F applied on A,

using free body diagram

For body m₁

F−T = m₁a ............(1)

for body m₂,

T = m₂a ................(2)

solving equation (1) and (2),

a = F (m₁+m₂)

a=600(10+20)

a=20m/s²

Now, 600−T=20∗20........................(Using eqn. (1))

T=200N

The free body diagram of 3 kg block is as shown in the fig. (a).
The equation of motion of 3 kg block is T₂ − 3g = 3a 
T₂ = 3(a + g) = 3(2 + 10) = 36N     ............(i)
The free body diagram of 5 kg is as shown in the Fig.(b).

The equation of motion of 5kg block is 
T₁ −T₂ − 5g = 5a 1
= 5(a + g) + T₂
= 5(2 + 10) + 36 = 96N         (Using (i))

The force acting down the plane for blocks A and B are:

Since F_A > F_B the block A moves down the plane.

If a is downward acceleration of 4 kg block, the upward acceleration of 1 kg block must be 2a.

If T is tension in each part of string, then equations of motion of the two blocks are
4a = 4g − 2T ...(i)
1 × 2a = T −1 g ..(ii)
or 4a = 2T −2g ...(iii)
Adding (i) and (iii), we get
8a  = 2g, a = g / 4
∴ Acceleration of 1 kg block = 2a = g/2 upwards.

Let a be the common acceleration of the system and T be tension of the string. The equations of motion of two blocks are
T − 8g = 8a …(i)
and 12g −T = 12a

Adding (i) and (ii), we get
4g = 20a  or a = g/5

Here, mass of monkey, m = 40kg
Maximum tension the rope can stand, T = 500N
Tension in the rope will be equal to apparent weight of the monkey (R).
The rope will break when R exceeds T.(a) When the monkey climbs up with an acceleration a = 5m s⁻²
R = m (g + a) = 40(10 + 5) = 600N
∴  R > T
Hence, the rope will break.
(b) When the monkey climbs down with an acceleration a=5m s⁻²
R = m (g − a) = 40(10 − 5) = 200N
∴ R < T
Hence, the rope will not break.
(c) When the monkey climbs up with a uniform speed v = 5m s⁻¹, its acceleration a = 0
∴ R = mg = 40 × 10 = 400N
∴ R < T
Hence, the rope will not break.
(d) When the monkey falls down the rope freely under gravity
a = g
∴ R = m(g − a) = m(g − g) = zero
Hence, the rope will not break.

Here, m₁ = 40kg

m₂ = 30kg, θ = 30^∘
Let T be the tension in the string and a be the acceleration of the system.
Their equations of motion are m₁gsin30^∘ −T = m₁a …(i)
T − m₂gsin30^∘ = m₂a …(ii)
Adding (i) and (ii), we get
(m₁ + m₂) a = (m₁ − m₂) gsin30^∘
Substituting the given values, we get
(40 + 30) a = (40 − 30) × 9.8 × 1/2 = 49
a = 49/70 = 0.7m s⁻²

The weight (W) and the action (A) both will act in same direction and parallel to each other. So the angle between them should be zero.