Question

A lighter block (3 kg) and a heavier block (12 kg) sit on a frictionless surface. Both blocks are initially at rest. The same force of 6 N then pushes to the right on each block for a distance of 16 m. What are the initial kinetic energies of the blocks? What are the changes in kinetic energy of the blocks? What are the final kinetic energies of the blocks? What are the initial momenta of the blocks? What are the final momenta of the blocks?

Answer 1

here,

for lighter block

mass , m = 3 kg

the force applied , F = 6 N

s = 16 m

the initial kinetic energy , KEi = 0.5 * m * u^2

KEi = 0.5 * m * 0^2 = 0 J

using Work-energy theorm

the change in kinetic energy , dKE = net Work done

dKe = F * s = 6 * 16 J = 96 J

let the final kinetic energy be KEf

we know , dKE = KEf - KEi

96 = KEf - 0

KEf = 96 J

the final kinetic energy is 96 J

the initial momentum of lighter block , Pi = m * u = 0 m/s

the final momentum of the lighter block , Pf = sqrt(2 * m * KEf )

Pf = sqrt(2 * 3 * 96) = 24 kg.m/s

for heavier block

mass , m' = 12 kg

the force applied , F = 6 N

s = 16 m

the initial kinetic energy , KEi' = 0.5 * m * u^2

KEi' = 0.5 * m * 0^2 = 0 J

using Work-energy theorm

the change in kinetic energy , dKE' = net Work done

dKE' = F * s = 6 * 16 J = 96 J

let the final kinetic energy be KEf'

we know , dKE' = KEf' - KEi'

96 = KEf' - 0

KEf' = 96 J

the final kinetic energy is 96 J

the initial momentum of lighter block , Pi' = m' * u = 0 m/s

the final momentum of the lighter block , Pf' = sqrt(2 * m' * KEf' )

Pf' = sqrt(2 * 12 * 96) = 48 kg.m/s