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Newton's second law of motion

A fighter plane is pulling out for a dive at a speed of 900 {~

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Solving time: 3 mins

A fighter plane is pulling out for a dive at a speed of $900 km / h$ . Assuming its path to be a vertical circle of radius $2000 m$ and its mass to be $16000 kg$ , find the force exerted by the air on it at the lowest point. Take, $g = 9.8 m / s^{2}$ .

Name: Video solution 1: A fighter plane is pulling out for a dive at a speed of 900 km/h. Assuming its path to be a vertical circle of radius 2000 m and its mass to be 16000 kg, find the force exerted by the air on it at the lowest point. Take, g=9.8 m/s2.
Uploaded: 2023-01-01T05:02:39.000Z
Duration: 2 min
Description: Video solution 1: A fighter plane is pulling out for a dive at a speed of 900 km/h. Assuming its path to be a vertical circle of radius 2000 m and its mass to be 16000 kg, find the force exerted by the air on it at the lowest point. Take, g=9.8 m/s2.

Views: 5,842 students

Updated on: Feb 19, 2023

Book:

Mechanics Volume 1 (DC Pandey)

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Sophia Discussed

A fighter plane is pulling out for a dive at a speed of

900 km / h

. Assuming its path to be a vertical circle of radius

2000 m

and its mass to be

16000 kg

, find the force exerted by the air on it at the lowest point. Take,

g = 9.8 m / s^{2}

7 mins ago

Discuss this question LIVE

7 mins ago

Text solutionVerified

At the lowest point in the path, the acceleration is vertically upward (towards the centre) and its magnitude is

v^{2} / r

.
The forces on the plane are :
(a) weight

M g

downward and
(b) force

F

by the air upward.
Hence, Newton's second law of motion gives

F - M g = M v^{2} / r or F = M (g + v^{2} / r)

Here,

v = 900 km / h = \frac{9 \times 1 0 ^{5}}{3600} m / s = 250 m / s

F = 16000 (9.8 + \frac{62500}{2000}) N = 6.56 \times 1 0^{5} N

(upward).

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Question Text	A fighter plane is pulling out for a dive at a speed of $900 km / h$ . Assuming its path to be a vertical circle of radius $2000 m$ and its mass to be $16000 kg$ , find the force exerted by the air on it at the lowest point. Take, $g = 9.8 m / s^{2}$ .
Updated On	Feb 19, 2023
Topic	Laws of motion
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 2
Upvotes	267
Avg. Video Duration	3 min

A fighter plane is pulling out for a dive at a speed of $900 km / h$ . Assuming its path to be a vertical circle of radius $2000 m$ and its mass to be $16000 kg$ , find the force exerted by the air on it at the lowest point. Take, $g = 9.8 m / s^{2}$ .

Text solutionVerified

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A fighter plane is pulling out for a dive at a speed of 900 km/h. Assuming its path to be a vertical circle of radius 2000 m and its mass to be 16000 kg, find the force exerted by the air on it at the lowest point. Take, g=9.8 m/s2.

Text solutionVerified

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A fighter plane is pulling out for a dive at a speed of $900 km / h$ . Assuming its path to be a vertical circle of radius $2000 m$ and its mass to be $16000 kg$ , find the force exerted by the air on it at the lowest point. Take, $g = 9.8 m / s^{2}$ .