Question

%3Cp%3EA%20drag%20force%20on%20a%20falling%20object%20in%20air%20is%20directly%20proportional%0Ato%20the%20square%20of%20its%20velocity.%20After%20falling%20some%20known%20distance%2C%0Athe%20kinetic%20energy%20of%20an%20object%20becomes%20K.%20After%20falling%20the%20same%0Adistance%20what%20will%20be%20the%20kinetic%20energy%20of%20an%20object%20that%20is%20n%0Atimes%20larger%3F%3C%2Fp%3E%0A%3Cp%3EAnswer%3A%20K%E2%80%99%3Dn4%20K%3C%2Fp%3E%0A

Answer 1

let the distance falled initally be d.

force is directly proportional to square of velocity.

F = l* (V^2)

l is some constant.

=>ma= l * (V^2)

and a = (V^2 - u^2)/2d

=> m*(V^2 - u^2)/2d = l*(V^2)

from this we get

mV^2 - mu^2 = 2dlV^2

=> V^2( m + 2dl) = mu^2

=> V^2 = mu^2/(m+2dl)

2dl is very large compared to nm

so V^2 = mu^2/2dl

also K = 0.5*m*(V^2)

now in the second case distance is same =d

initial velocity is same u

l is constant

m = nm

so V^2 = nm*u^2/(2dl)

=> V(in second case ) = 2V(in first case)

so now Kinetic energy = 0.5 *nm*(2V)^2

=> 4nK