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

Answer:

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

Explanation:

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

$K.E = \dfrac{1}{2}mv^2$

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

$K.E = \dfrac{1}{2}(24kg)(12.1)^2\\\\K.E = 1756.92J$

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,

$\dfrac{1}{2}mv_t^2= 388.92J$

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.