1.

Explain work energy theorem .

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 .


Discussion

No Comment Found