1.

A person strikes a nail with a hammer of mass `500g` moving with a velocity of `10m//s`. The hammer comes to rest in `0.01 s` after striking the nail. Calculate (i) force exerted by the hammer on the nail (ii) distance moved by the nail into the plank.

Answer» Here, `m=500g= 0.5 kg, u=10m//s, v-0, t=0.01 s`.
From `v= u+at`
`0=10+axx0.01, a= (-10)/(0.01)= - 1000m//s^(2)`
Force exerted by hammer
`F= ma= 0.5xx(-1000)= -500N`
From `v^(2)-u^(2)= 2as`
`s=(v^(2)-u^(2))/(2a)= (0-10^(2))/(2(-1000))= 0,05m`


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