Question

a block of mass m is projected vertically upwards with velocity v reaches ground with speed 0.6 v. work done by resistive forces acting on the block is​

Answer 1

hence, workdone by the resistive forces acting on the block is 0.32 mv².

mass of block is m

initial velocity of block is v

so, initial kinetic energy of block, K.E_i = 1/2 mv²

final velocity of block is 0.6v

so, final kinetic energy, K.E_f = 1/2 m(0.6v)² = 1/2 mv²(0.36)

from conservation of work - energy theorem,

workdone by the resistive forces acting on the block = lost in kinetic energy

= 1/2 mv² - 1/2 mv²(0.36)

= 1/2 mv²(1 - 0.36)

= 1/2 mv²(0.64)

= 0.32 mv²

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Answer 2

$\huge\bold\purple{Answer:-}$

mass of block is m

initial velocity of block is v

so, initial kinetic energy of block, K.E_i = 1/2 mv²

final velocity of block is 0.6v

so, final kinetic energy, K.E_f = 1/2 m(0.6v)² = 1/2 mv²(0.36)

from conservation of work - energy theorem,

workdone by the resistive forces acting on the block = lost in kinetic energy

= 1/2 mv² - 1/2 mv²(0.36)

= 1/2 mv²(1 - 0.36)

= 1/2 mv²(0.64)

= 0.32 mv²