Question

A body of mass 10kg at rest is acted upon simultaneous by two forces 4n and 3n at right angles to each other. The ke of the body at the end of 10 sec is

Answer 1

Answer:

Notice that the numbers (3 and 4) happen to be part of a Pythagorean triple (3–4–5), which lazy physics teaches love to use in problems because it makes grading simpler!

Orthogonal vectors of magnitude 3N and 4N, acting at right angles, resolve to a composite vector whose magnitude is that of the hypotenuse of the right-triangle that they form.

If a force of 5N acts upon a mass of 2 Kg, then (considering Newton’s Second Law: F=ma, or a=F/m), the acceleration will be 5N/2Kg.

Since one Newton is one Kilogram-meter-per-second-squared, 5N/2Kg becomes: (5 Kg m / s^2) / (2Kg), which simplifies to: (5 m / s^2) / (2) or (5/2) m/s^2.

                                                            or

With the angle is 90°, the resultant is 5 N.

By using F = m x a

5 = 2 x a

Then,

a = 5/2 = 2.5 m/s²

The detailed answer is above. A simplified answer is below.

Explanation:

Answer 2

Since it is acted upon simultaneously by two forces 4N and 3N at right angles, net force =

$\sqrt{ {4}^{2} + {3}^{2} } = 5n$

The mass is given as 10Kg→a=Fm10Kg→a=Fm=510=510=0.5m/s2=0.5m/s2

Now, we can calculate the Kinetic Energy as K.E=12K.E=12mv2=12mv2=12m(at)2m(at)2

⇒KE=12⇒KE=1210×0.52×10s10×0.52×10s=125J