Question

3 A jet fighter has a mass at take-off of 28 000 kg.

a)Its engines produce a force of 70 000 N. What
is its acceleration?
b) What is its acceleration if the force from the
engine is increased to 120 000 N?

Answer 1

A. 2.5 m/s²

B. 4.29 m/s²

Step-by-step explanation:

A. Determination of the acceleration.

Mass (m) = 28000 Kg

Force (F) = 70000 N

Acceleration (a) =?

Force (F) is simply defined as the product of mass (m) and acceleration (a). Mathematically, it is expressed as:

Force (F) = mass (m) × acceleration (a)

F = ma

Divide both side by m

a = F / m

With the above formula, we can obtain the acceleration as follow:

Mass (m) = 28000 Kg

Force (F) = 70000 N

Acceleration (a) =?

a = F / m

a = 70000 /28000

a = 2.5 m/s²

B. Determination of the acceleration.

Mass (m) = 28000 Kg

Force (F) = 120000 N

Acceleration (a) =?

a = F / m

a = 120000 /28000

a = 4.29 m/s²