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Given KE=(1)/(2)mv^(2). If m=10kg, v=5m/...

Given `KE=(1)/(2)mv^(2)`. If `m=10kg, v=5m//s` then find the value of KE?

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The correct Answer is:
125 J
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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

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") Two bodies of mass 1 kg and 4 kg possess equal momentum. The ratio of their kinetic energies is

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") Two bodies masses 1 kg and 4 kg having equal kinetic energies. The ratio of their momentum is

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") Two bodies A and B of unequal masses having same momentum have masses in the ratio 1 : 2 then their K.E are in the ratio

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") What will be the momentum of a body of mass 100 g having kinetic energy of 20 J ?

Moment of inertia of a body about a given axis is the rotational inertia of the body about that axis. It is respresented by I = MK^(2) , where M is mass of body and K is radius of gyration of the body about that axis. It is a scalar quantity, which is measured in kg m^(2) . when a body rotates about a given axis, and teh axis of rotates also moves, then total K.E. of body = K.E. of translation + K.E. of rotation E = (1)/(2)mv^(2) + (1)/(2)I omega^(2) Which the help of the compreshension given above, choose the most apporpriate altermative for each of the following questions : Kinetic energy of rotation of the flywheel in the above case is

Moment of inertia of a body about a given axis is the rotational inertia of the body about that axis. It is respresented by I = MK^(2) , where M is mass of body and K is radius of gyration of the body about that axis. It is a scalar quantity, which is measured in kg m^(2) . when a body rotates about a given axis, and teh axis of rotates also moves, then total K.E. of body = K.E. of translation + K.E. of rotation E = (1)/(2)mv^(2) + (1)/(2)I omega^(2) Which the help of the compreshension given above, choose the most apporpriate altermative for each of the following questions : A 40 kg flywheel in the from of a unifrom circular disc of diameter 1 m is making 120 rpm . Its moment of inertia about a transverse axis through its centre is

If in the previous problem , the final K.E. is 3//5 of the initial K.E. , find the value of e .

If the acceleration of the block B in the following system is a ( in m//s^(2)) then find out value of 2a//5(g=10m//s^(2)) :

A block of mass 2kg is sliding on a smooth surface. At t=0, its speed is v_(0)=2m//s . At t=0, a time varying force starts acting on the block in the direction opposite to v_(0) if the speed of object is (38)/(N)m//s at t=10s. Find the value of N

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