Question

My Notes You push a box of mass 24 kg with your car up to an icy hill slope of irregular shape to a height 5.7 m. The box has a speed 12.1 m/s when it starts up the hill, the same time that you brake. It then rises up to the top (with no friction) with a horizontal velocity before immediately falling off a sheer cliff to the ground (with no drag). (a) What is the speed of the box at the top of the hill?

Answer 1

The speed of the box at the top of the hill will be 5.693m/s.

Step-by-step explanation:

The kinetic energy of the box at the bottom of the hill is

`K.E = (1)/(2)mv^2`

putting in
`m =24kg` and
`v = 12.1m/s` we get

`K.E = (1)/(2)(24kg)(12.1)^2\\\\K.E = 1756.92J`

Now, the potential energy this box gains as it rises
`h =5.7m` up the hill is

`P.E = mgh`

`P.E = (24kg)(10ms/s^2)(5.7m)\\\\P.E = 1368`

Therefore, the energy left
`E_(left)` in the box at the top if the hill will be

`E_(left) =K.E - P.E = 1756.92J-1368J\\`

`\boxed{E_(left) = 388.92J}`

This left-over energy must appear as the kinetic energy of the box at the top of the hill (where else could it go? ); therefore,

`(1)/(2)mv_t^2= 388.92J`

putting in numbers and solving for
`v_t` we get:

`\boxed{v_t = 5.693m/s.}`

Thus, the speed of the box at the top of the hill is 5.693m/s.