" Define kinetic energy.Prove that "KE=1/2mv2" (mv square)."

Kinetic energy of a body (KE) =

Relation between kinetic energy and momentum Let us consider a body of mass 'm' having a velocity 'v', then momentum of the body P = mass xx velocity P = m xx v rArr v = (P)/(m) " "...(1) From definition, kinetic energy (K.E) of the body K.E = (1)/(2) mv^(2)" "...(2) Now putting the value of (1) in (2) we have K.E = (1)/(2)m ((P)/(m))^(2) K.E. = (1)/(2)m (P^(2))/(m^(2)) = (1)/(2) (P^(2))/(m) = (P^(2))/(2m)" "...(3) Thus we can write P^(2) = 2m xx K.E rArr P = sqrt(2m xx K.E) Thus momentum = sqrt(2 xx "mass" xx "kinetic energy") The kinetic energy of a given body is doubled. Its momentum will

The percentage errors in the measurements of mass and speed are 1% and 2% respectively. What is the % error in the kinetic energy (K.E = 1/2mv^2) ?

A body whose mass is 3kg performs rectilinear motion according to the formula s = 1 + t + t^2 , where s is measured the kinetic energy 1/2 mv^2 and t in second. Determine the kinetic energy 1/2 mv^2 of the body in 5 sec after its start.

A solid sphere is moving on a horizontal plane. Ratio of its translational K.E. and rotational kinetic energy is [Hint : E_L/E_r = (1/2mv^2)/(1/2Iomega^2) = (mv^2)/(2/5mr^2 omega^2) = (mv^2)/(2/5mv^2) = 5/2 ]

If v=(3hati+2hatj+6k) m/s and m=(2)/(7)kg then find kinetic energy (i.e. (1)/(2)mv^(2))

If v=(3hati+2hatj+6k) m/s and m=(2)/(7)kg then find kinetic energy (i.e. (1)/(2)mv^(2))

If vecv=(3hati+2hatj+6k) m/s and m=(2)/(7)kg then find kinetic energy (i.e. (1)/(2)mv^(2))