Question

An automobile having a mass of 1100 kg initially moves along a level highway at 110 km/h relative to the highway. It then climbs a hill whose crest is 40 m above the level highway and parks at a rest area located there. Use a reference with kinetic and potential energy each equal to zero for the stationary highway before the hill.

Let g = 9.81 m/s2.
For the automobile, determine its change in kinetic energy and its change in potential energy, both in kJ.

Answer 1

Change in kinetic energy=-513.652 KJ

Change in potential energy=431.64KJ

Step-by-step explanation:

We are given that

Mass of an automobile , m=1100 kg

Initial speed, u=110 km/h=
`110* (5)/(18)=30.56 m/s`

Where
`1 km/h=5/18 m/s`

Height ,
`h_2=40 m`

`g=9.81 m/s^2`

Final speed, v=0

Change in kinetic energy,
`\Delta K.E=(1)/(2)m(v^2-u^2)`

`\Delta K.E=(1)/(2)(1100)(0-(30.56)^2)=-513652.48 J`

`\Delta K.E=-(513652.48)/(1000)=-513.652 KJ`

Where 1 KJ=1000 J

Change in potential energy,
`\Delta P.E=mgh(h_2-h_1)`

Initially height, h1=0

Using the formula

`\Delta P.E=1100* 9.81(40-0)`

`\Delta P.E=431640J`

`\Delta P.E=431.64KJ`