Question

An aeroplane of mass 10000 kg required a speed of 20 ms for take-off, the run on the ground being 100m. The coefficient of Kinetic friction between the wheels of the plane and ground is 0.3 Assume the plane accelerates uniformly during the take-off Determine the minimum force required by the engine of the plane to take off (g=10ms).

Answer 1

Answer:-

Given:-

m=10,000kg.

m=10,000kg.v=20 m/s.

m=10,000kg.v=20 m/s.s=100m.

m=10,000kg.v=20 m/s.s=100m. coefficient of Kinetic friction =0.3

Explanation:-

As the plane is going to taking off the initial velocity of plane is zero.

From third kinematic equation.

${v}^{2} - {u}^{2} = 2as$

u=0.

${20}^{2} = 2 \times 100 \times a$

$\frac{400}{200} = a$

$a = 2 \frac{m}{ {s}^{2} }$

so, acceleration of plane is 2m/s^2.

Now,

$weight = mg = normal \: reaction$

$w = 10000 \times 10 = {10}^{5} newton$

frictional force is =coefficient of Kinetic friction× Normal reaction.

$f = {10}^{5} \times 0.3$

$f = 3 \times {10}^{4} \: newton$

Now taking F.B.D of aeroplane

#refer the attachment for the figure.

$F-f=ma \:$

$F-30000=10000 \times 2$

$F = 20000 + 30000$

$F = 50000 \: newton$

$F = 5 \times {10}^{4} newton$

or

$F = 50 \: kilo \: newtons$

so,the force applied by engiene should be 50 kilo Newton to take off by opposing friction.