Q#13
The elevator shown in figure (5-E5) is descending with an acceleration of 2 m/s². The mass of the block A is 0.5 kg. What force is exerted by the block A on the block B?
Answer:
If the elevator was not moving the force applied by the block A on the block B would have been equal to its weight = mg = 0.5 kg x 9.8 m/s² = 4.9 N. But in this case the elevator is descending with an acceleration of a = 2 m/s².
So, in order to apply the Newton's Laws of motion with respect to the elevator we apply a pseudo force equal to ma on the block A in the direction opposite to the acceleration ie upwards. The net force on the block is
= mg - ma= 4.9 - 0.5 x 2 ≈ 4.0 N.
Q#14
A pendulum bob of mass 50 g is suspended from the ceiling of an elevator. Find the tension in the string if the elevator (a) goes up with acceleration 1.2 m/s², (b) goes up with deceleration 1.2 m/s², (c) goes up with uniform velocity, (d) goes down with acceleration 1.2 m/s², (e) goes down with deceleration 1.2 m/s² and (f) goes down with uniform velocity.
Answer:
Mass of the pendulum bob = 50 g = 0.05 kg.
Weight of the bob = 0.05 kg x 9.8 m/s² = 0.49 N (Downward)
It is also the tension in the string when the elevator is at rest.
When the elevator moves with an acceleration a we have to apply a pseudo force equal to ma on the bob opposite to the direction of a. So let us take the different situations taking downward direction as +ve :
(a) Acceleration a = -1.2 m/s² (Upwards),
Pseudo force = ma = 0.05 kg x 1.2 m/s² = 0.06 N (Downwards)
Weight = 0.49 N (Downwards)
Total force on the bob = 0.49 + 0.06 N = 0.55 N (Downwards)
Tension in the string = 0.55 N
(b) Deceleration upwards = 1.2 m/s²
Acceleration a = 1.2 m/s² (Downward),
Pseudo force = ma = -0.05 kg x 1.2 m/s² = -0.06 N (Upward)
Weight = 0.49 N (Downward)
Total force on the bob = 0.49 - 0.06 N = 0.43 N (Downwards)
Tension in the string = 0.43 N
(c) Elevator goes up with uniform velocity,
Acceleration a = 0 m/s²,
Pseudo force = ma = 0.05 kg x 0 m/s² = 0
Weight = 0.49 N (Downward)
Total force on the bob = 0.49 N (Downwards)
Tension in the string = 0.49 N
(d) Acceleration a = 1.2 m/s² (Downwards),
Pseudo force = ma = -0.05 kg x 1.2 m/s² = -0.06 N (Upwards)
Weight = 0.49 N (Downwards)
Total force on the bob = 0.49 - 0.06 N = 0.43 N (Downwards)
Tension in the string = 0.43 N
(e) Deceleration downwards = 1.2 m/s²
Acceleration a = -1.2 m/s² (Upwnward),
Pseudo force = ma = 0.05 kg x 1.2 m/s² = 0.06 N (Downward)
Weight = 0.49 N (Downward)
Total force on the bob = 0.49 + 0.06 N = 0.55 N (Downwards)
Tension in the string = 0.55 N
(f) Elevator goes down with uniform velocity,
Acceleration a = 0 m/s²,
Pseudo force = ma = 0.05 kg x 0 m/s² = 0 N
Weight = 0.49 N (Downward)
Total force on the bob = 0.49 N (Downwards)
Tension in the string = 0.49 N
Q#15
A person is standing on a weighing machine placed on the floor of an elevator. The elevator starts going up with some acceleration, moves with uniform velocity for a while and finally decelerates to stop. The maximum and minimum weights recorded are 72 kg and 60 kg. Assuming that the magnitude of acceleration and deceleration are the same, find (a) true weight of the person and (b) the magnitude of the acceleration. Take g = 9.9 m/s².
Answer:
Let the true weight of the person be 'm' kg and the acceleration of the elevator = 'a' m/s². When the elevator starts going up with acceleration, the pseudo force acts downwards which adds to the apparent weight of the person. In this case apparent weight = mg + ma = m(g + a).
But the recorded apparent weight = 72 kg, which gives the weight force = 72g N. These two should be equal.
So m(g + a) = 72g
mg + ma = 72g
During uniform velocity of elevator the weight does not change. When the elevator decelerates its acceleration is downwards and the pseudo force is upwards. So, the apparent weight = mg – ma which is recorded as 60 kg. In terms of weight-force these two can be equated as
mg - ma = 60 g (2)
Adding (1) and (2) we get
2 mg = 132 g
m = 132/2 = 66 kg.
So true weight of the person is 66 kg.
Subtract (B) from (A)
2 ma = 72 g - 60 g = 12 g
2 x 66 a = 12 x 9.9
a = 6 x 9.9/66 = 0.9 m/s².
So, acceleration of the elevator = 0.9 m/s².
Q#16
Find the reading of the spring balance shown in figure (5-E6). The elevator is going up with an acceleration of g/10, the pulley and the string are light and the pulley is smooth.
Since the elevator goes up with acceleration g/10 so the pseudo forces due to this acceleration on both the blocks will be downwards and equal to mass times this acceleration.
Let the accelerations of the blocks be 'a' with respect to the elevator (Upwards for left block and downwards for the right block). Let tension in the string be T.
Considering the forces and acceleration of the left block, we have,
T – 1.5g – 1.5g/10 = 1.5 a (1)
Similarly for the right block.
3g + 3g/10 - T = 3a
Adding the two equations we get,
1.5g + 1.5g/10 = 4.5a
a = g/3 + g/30 = 11g/30
Putting this value in (1),
T = 1.5 x 11g/10 + 1.5 x 11g/30 =16.5 x 4g/30 = 66g/30 = 11g/5 N
The pull on the spring balance = 2T = 2 x 11g/5 = 22g/5 = 4.4g N
The scale on the spring balance reads in kg, so its reading will be
= 4.4g/g = 4.4 kg.
Q#17
A block of 2 kg is suspended from the ceiling through a mass-less spring of spring constant k = 100 N/m. What is the elongation of the spring? If another 1 kg is added to the block, what would be the further elongation ?
Answer:
Weight force of the block = 2g N = 2 x 9.8 N = 19.6 N.
Elongation of the spring = 19.6/k = 19.6/100 m = 0.196 m ≈ 0.2 m
When another 1 kg block is added, total weight = 3 kg
Now weight force = 3g = 3 x 9.8 N = 29.4 N
Total elongation = 29.4/100 = 0.294 m ≈ 0.3 m
So further elongation = 0.3 - 0.2 m = 0.1 m
Post a Comment for "Solutions to Exercises on Newton's Laws of Motion HC Verma's Concepts of Physics Part 1 (13-17)"