Question

Please solve this question??

i use conservaion of energy bt still cant solve...

Answer 1

Option (d)

Explanation:

In position A

total energy `E_A=(KE)_A+(PE)_A=(60-20)J=40J`

In position B

total energy `E_B=(KE)_B+(PE)_B=(100+40)=140J`

Then in moving particle from `Ato B` the difference in total energy becomes `DeltaE=E_B-E_A=(140-40)J=100J`

So net work done by all forces in moving from `AtoB` will be `100J` and hence option (d) is accepted.