Question #993d6

Each in turn

Qu 15 . If we use the formula `v^2=u^2+2as`
Then since v, final velocity, is nil we can simplify to:
`-u^2=2as`

Using Newton's 2nd law ,`F=ma`, we can substitute for `a`
`-u^2=-2(F/m)s`
The negative values indicates that the force is decelerating and acts in the opposite direction to u.

Since we want to consider distance, `s`, we can rearrange:
`s=(u^2m)/(2F)`

For the 4kg mass travelling at 2m/s
`s=(2^2*4)/(2F)=16/(2F)`

For the 1kg mass travelling at 4m/s
`s=(4^2*1)/(2F)=16/(2F)`

Since F is the same, we can see that they take the same distance to stop.

We could also have used energy arguments that kinetic energy must equal work done to bring to a stop:
Hence `1/2mv^2=F.s`
Since the KE of the objects is the same, the distance moved by the braking force would be the same

Qu16
We need to consider work done against gravity (since there is no friction to do work against). The distance moved in the vertical direction is given by `0.5sin30^0=0.25m`

The weight is 10N (`1kg times 10ms^-2`)
So work done against gravity is:
`WD=F.s=10 times0.25=2.5J`

Qu 17 Both people are at the same height so have the same GPE (U) since this is a function of the height above the ground

Qu18

The rollercoaster has GPE and Kinetic energy

`=mgh +1/2mv^2`
`=(80times10times200) +(1.2times 80 times 20^2)`
`=176,000J=17.6times10^4J`