An iron nail is dropped from a height h on the level of a sand bed. If it penetrates through a distance x in the sand before coming to rest, the average force exerted by the sand on the nail is

Options

A. `mg((h)/(x)+1)`
B. `mg((x)/(h)+1)`
C. `mg((h)/(x)-1)`
D. `mg((x)/(h)-1)`

Answer

A

Explanation

The nail hits the sand with a speed `v_(0)` after falling a height h
`rArr v_(a)^(2)=2gh rArr v_(0)=sqrt(2gh) " " ...(1)`
The nail stops after sometime say, t penetrating through a distance, x into the sand. Since its velocity decreases gradually the sand exerts a retarding upward force, R (say).

The net force acting on the nail is given as
`SigmaF_(y)=(mg-R)= -ma`
`rArr R=m(g+a)" "...(2)`
Where a = deceleration of the nail. Since the nail penetrates a distance x
`0-v_(0)^(2)= -2ax " " ...(3)`
Putting `v_(0)` from (1) and .a. from (2) in (3), we obtain
`2gh=2((R-mg)/(m)) x rArr R =(mg(h+x))/(x)`
` rArr R=mg((h)/(x)+1)`