Newton’s Second Law corrected the old belief that force is needed to maintain motion. This idea came from the philosopher ______.
Advertisements
Advertisements
Question
A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s–2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the
(a) force on the floor by the crew and passengers,
(b) action of the rotor of the helicopter on the surrounding air,
(c) force on the helicopter due to the surrounding air.
Advertisements
Solution 1
(a) Mass of the helicopter, mh = 1000 kg
Mass of the crew and passengers, mp = 300 kg
Total mass of the system, m = 1300 kg
Acceleration of the helicopter, a = 15 m/s2
Using Newton’s second law of motion, the reaction force R, on the system by the floor can be calculated as:
R – mpg = ma
= mp(g + a)
= 300 (10 + 15) = 300 × 25
= 7500 N
Since the helicopter is accelerating vertically upward, the reaction force will also be directed upward. Therefore, as per Newton’s third law of motion, the force on the floor by the crew and passengers is 7500 N, directed downward.
(b) Using Newton’s second law of motion, the reaction force R’, experienced by the helicopter can be calculated as:
R' - mg = ma
=m(g+a)
=1300(10+15) = 1300 xx 25
= 32500 N
The reaction force experienced by the helicopter from the surrounding air is acting upward. Hence, as per Newton’s third law of motion, the action of the rotor on the surrounding air will be 32500 N, directed downward.
(c) The force on the helicopter due to the surrounding air is 32500 N, directed upward.
Solution 2
Here, mass of helicopter, m1= 1000 kg
Mass of the crew and passengers, m2 = 300 kg upward acceleration, a = 15 ms-2and g = 10 ms-2
(a)Force on the floor of helicopter by the crew and passengers = apparent weight of crew and passengers
= m2 (g + a) = 300 (10 + 15) N = 7500 N
(b)Action of rotor of helicopter on surrounding air is obviously vertically downwards, because helicopter rises on account of reaction to this force. Thus, force of action
F = (m1+ m2) (g + a) = (1000 + 300) (10 + 15) = 1300 x 25 = 32500 N
(c)Force on the helicopter due to surrounding air is the reaction. As action and reaction are equal and opposite, therefore, force of reaction, F = 32500 N, vertically upwards.
RELATED QUESTIONS
A body of mass 0.40 kg moving initially with a constant speed of 10 m s–1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = –5 s, 25 s, 100 s.
A smooth wedge A is fitted in a chamber hanging from a fixed ceiling near the earth's surface. A block B placed at the top of the wedge takes time T to slide down the length of the wedge. If the block is placed at the top of the wedge and the cable supporting the chamber is broken at the same instant, the block will.
A free 238U nucleus kept in a train emits an alpha particle. When the train is stationary, a nucleus decays and a passenger measures that the separation between the alpha particle and the recoiling nucleus becomes x at time t after the decay. If the decay takes place while the train is moving at a uniform velocity v, the distance between the alpha particle and the recoiling nucleus at a time t after the decay, as measured by the passenger, is
If the tension in the cable supporting an elevator is equal to the weight of the elevator, the elevator may be
(a) going up with increasing speed
(b) going down with increasing speed
(c) going up with uniform speed
(d) going down with uniform speed
A particle of mass 0.3 kg is subjected to a force F = −kx with k = 15 N/m. What will be its initial acceleration if it is released from a point x = 20 cm?
A small block B is placed on another block A of mass 5 kg and length 20 cm. Initially, the block B is near the right end of block A (In the following Figure). A constant horizontal force of 10 N is applied to the block A. All the surfaces are assumed frictionless. Find the time that elapses before block B separates from A.

Let m1 = 1 kg, m2 = 2 kg and m3 = 3 kg in the following figure. Find the accelerations of m1, m2 and m3. The string from the upper pulley to m1 is 20 cm when the system is released from rest. How long will it take before m1 strikes the pulley?

A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of 12 m/s2. Find the displacement of the block during the first 0.2 s after the start. Take g = 10 m/s2.
State Newton's second law of motion. Under what condition does it take the form F = ma?
A pebble is thrown vertically upwards with a speed of 20 m s-1. How high will it be after 2 s? (Take g = 10 m s-2)
A pebble is dropped freely in a well from its top. It takes 20 s for the pebble to reach the water surface in the well. Taking g = 10 m s-2 and speed of sound = 330 m s-1. Find : The time when echo is heard after the pebble is dropped.
A motorcycle of mass 100 kg is running at 10 ms−1. If its engine develops an extra linear momentum of 2000 Ns, calculate the new velocity of a motorcycle.
Multiple Choice Question. Select the correct option.
The impulse of a body is equal to:
What causes motion in a body?
State Newton's second law of motion. Is Newton's first law of motion contained in Newton's second law of motion?
A stone is dropped from a cliff 98 m high.
What will be its speed when it strikes the ground?
A stone is dropped from a tower 98 m high. With what speed should a second stone be thrown 1 s later so that both hit the ground at the same time?
The position time graph of a body of mass 2 kg is as given in figure. What is the impulse on the body at t = 0 s and t = 4 s.

Figure shows (x, t), (y, t ) diagram of a particle moving in 2-dimensions.
|
|
![]() (b) |
If the particle has a mass of 500 g, find the force (direction and magnitude) acting on the particle.


