State and derive work-energy relationship.ORDerive a relationship betw

1.	State and derive work-energy relationship.ORDerive a relationship between kinetic energy and work.
Answer» it states that change in kinetic energy of a body is equal to work done and vice versa. Let a constant force \(\vec{F}\) be applied to a body moving with initial velocity \(\vec{u},\) so that its velocity becomes \(\vec{v}\) along the direction of force when S is its displacement. Using Newton’s second law of motion we get magnitude of force F = ma and from equation of motion, we get v² − u² = 2as, where a is the acceleration of the body. Multiplying both sides by m/2, we get \(\frac{1}{2}mv^2-\frac{1}{2}mu^2=maS\) i.e., \(\frac{1}{2}mv^2-\frac{1}{2}mu^2=FS=W\) i.e., K.E._(f) is final kinetic energy and K.E._(i) is initial kinetic energy. Thus work done on a body by a net force is equal to the change in kinetic energy of the body.

State and derive work-energy relationship.ORDerive a relationship between kinetic energy and work.

Answer»

it states that change in kinetic energy of a body is equal to work done and vice versa. Let a constant force \(\vec{F}\) be applied to a body moving with initial velocity \(\vec{u},\) so that its velocity becomes \(\vec{v}\) along the direction of force when S is its displacement. Using Newton’s second law of motion we get magnitude of force F = ma and from equation of motion, we get v² − u² = 2as, where a is the acceleration of the body.

Multiplying both sides by m/2, we get

\(\frac{1}{2}mv^2-\frac{1}{2}mu^2=maS\)

i.e., \(\frac{1}{2}mv^2-\frac{1}{2}mu^2=FS=W\)

i.e., K.E._(f) is final kinetic energy and K.E._(i) is initial kinetic energy.

Thus work done on a body by a net force is equal to the change in kinetic energy of the body.

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