Question

A wooden box of mass 5.0kg is released to fall through a height of 20m to hit the ground.the work done against resistance is 300J . Find the speed of the box before hitting the ground?

Answer 1

`16.5ms^-1`, rounded to one decimal place

Explanation:

Work done against resistance`=300J`
Using the following expression to find the force of resistance `vecF`
`"Work done"=vecFcdot vecs`

Inserting given values we get
`300=vecFxx20 cos theta`
`theta=180, => cos180^@=-1`
`=>vecF=-300/20=-15N`

To calculate deceleration due to resistance we use Newton's Second Law of motion
`vecF=mveca`
Inserting given values we get
`-15=5veca`
`=>veca=-15/5=-3ms^-2`

Net downwards acceleration of the wooden box under free fall
`vecg-veca=9.81-3=6.81ms^-2`

Using the kinematic expression
`v^2-u^2=2as`
We get
`v^2-0^2=2xx6.81xx20`
`=>v=sqrt(2xx6.81xx20)`
`=>v=16.5ms^-1`, rounded to one decimal place
.-.-.-.-.-.-.--.

Alternative method

Initial Potential energy`=mgh=5.0xx9.81xx20=981J`
Initial Kinetic energy`=1/2m u^2=1/2xx5.0xx0^2=0J`
Initial Total energy `=981+0=981J`

Work done against resistance`=300J`
Using Law of conservation of energy
Final Total energy`=981-300=681J` ....(1)

Final Potential energy`=mgh=5.0xx9.81xx0=0J`
Final Kinetic energy`=1/2m v^2=1/2xx5.0xxv^2=2.5v^2J`
Final total energy `=0+2.5v^2=2.5v^2J` ......(2)

Equating Final total energy obtained in (1) and (2) we get
`2.5v^2=681`
`=>v^2=681/2.5`
`=>v=sqrt(681/2.5)`
`=>v=16.5ms^-1`, rounded to one decimal place