During a rockslide, a 670 kg rock slides from rest down a hillside that is 740 m along the slope and 240 m high. The coefficient of kinetic friction between the rock and the hill surface is 0.25. (a) If the gravitational potential energy U of the rock-Earth system is zero at the bottom of the hill, what is the value of U just before the slide? (b) How much energy is transferred to thermal energy during the slide? (c) What is the kinetic energy of the rock as it reaches the bottom of the hill? (d) What is its speed then?

a)

b)

c)

d) 35.8 m/s

Explanation:

a)

The gravitational potential energy of an object is the energy possessed by the object due to its location with respect to the ground.

It is given by:

where

m is the mass of the object

g is the acceleration due to gravity

h is the height of the object, relative to a reference level

Here, the reference level is taken at the bottom of the hill (where the potential energy is zero).

So, we have:

m = 670 kg is the mass of the rock

h = 240 m is the initial height of the rock

So, the potential energy of the rock just before the slide is

b)

The energy transferred to thermal energy during the slide is equal to the work done by friction, which is:

where

is the force of friction

d = 740 m is the displacement of the rock along the ramp

The force of friction is given by:

where

is the coefficient of friction

m = 670 kg is the mass of the rock

is the angle of the ramp

Since we know the lenght of the ramp (d = 740 m) and the height (h = 240 m), we can find the angle:

Therefore, the work done by friction is:

So, the energy transferred to thermal energy is .

c)

According to the law of conservation of energy, the kinetic energy of the rock as it reaches the bottom of the hill will be equal to the initial potential energy (at the top) minus the energy transformed into thermal energy.

Therefore, we have:

where here we have:

is the potential energy of the rock at the top of the hill

is the energy converted into thermal energy

Substituting, we find

So, this is the kinetic energy of the rock at the bottom of the hill.

d)

The kinetic energy of the rock at the bottom of the hill can be rewritten as

where

m is the mass of the rock

v is its final speed

In this problem, we have:

is the final kinetic energy of the hill

m = 670 kg is the mass of the rock

Therefore, the final speed of the rock is:

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