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Explain work energy theorem . |
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Answer» Solution :For straightline motion : If body is MOVING with initial velocity u and constantacceleration a , it acquired velocity v with displacement s THENFROM the third equationof motion , `v^(2) -u^(2) = 2as` multiplying `m/2 `on both sides of equation , `1/2 mv^(2) -1/2 "mu"^(2) =mas ` `K_(f) -K_(i)=Fs [ :. ma =F] ` For motion in three dimension : `v^(2) -u^(2) = 2vec(a) .vec(d)` where v = final velocity , u = INTIAL velocity , `vec(a)` = constant acceleration and `vec(d)` = displacementMultiplying by `m/2 ` on both side of equation , ` 1/2 mv^(2) -1/2 "mu"^(2) = m vec(a) .vec(d)` `vec(F) .vec(d) ""[ :. mvec(a) = vec(F)] ` The left hand side is the change in kinetic energy of body , whileright HANDIS the multiplication of force and the displacement .It is known as work and it is denoted by W. ` :. K_(f) - K_(i) = W and DeltaK = W ` This RELATION is known as work energy theorem . Note : If a body undergoes a displacement in the direction of net force , then work done on the body which is positive andthe speed and kineticenergy of body increases but if force and displacement are in opposite direction , then work done by the body which is negative and the speed and kinetic energy decreases . |
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