Question

A body of mass 5 kg is lifted vertically at a constant velocity of 12m. calculate a) the force applied b) work done in lifting the body c) what happens to the work performed?

Answer 1

a. acc to Newton's 2nd law of motion,
F=ma.
therefore, force= 5x 9.8 ( 9.8 is the acceleration due to gravity)
f = 49N
B. w= fxs (s is the displacement)
w= kinetic energy.
therefore, k.e. = 1/2mv²
k.e. is equal to 1/2x 5x 12x12
= 360 joule.
therefore work done is 3.6 x 10² joule.

Answer 2

Force, F = 49 N

Work done, W = 360 J

Explanation:

Given that,

Mass of the body, m = 5 kg

It is lifted vertically at a constant velocity of 12 m/s

(a) The force acting on the body is called gravitational force. The force applied is given by :

F = mg

$F = 5\ kg\times 9.8\ m/s^2\\\\F=49\ N$

(b) The work done in lifting the body is given by :

$W=\dfrac{1}{2}mv^2\\\\W=\dfrac{1}{2}\times 5\times 12^2\\\\W=360\ J$

(c) The work performed is stored in the form of energy.

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Work done

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