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Conservation of Energy Formula: A Rock Falling from a 20 m Cliff

Under the textbook’s frictionless model, the falling rock retains 980 J of potential energy and has gained 980 J of kinetic energy after falling halfway.

Question

A 10 kg rock falls from a 20 m cliff. What is the kinetic and potential energy when the rock has fallen 10 m?

Answer

The textbook’s worked solution lists v1=0v_1=0 and g=9.80 m/s2g=9.80\,\mathrm{m/s^2}, and the section uses a frictionless approximation for these mechanical-energy transformations. Under that textbook model, mechanical energy is conserved:

KEi+PEi=KEf+PEf.KE_i+PE_i=KE_f+PE_f.

Using the textbook’s v1=0v_1=0, KEi=0KE_i=0. Its initial potential energy is

PEi=mghi=(10)(9.80)(20)=1960 J.PE_i=mgh_i=(10)(9.80)(20)=1960\,\text{J}.

After falling 10 m, the rock is still 10 m above the ground. Its potential energy is

PEf=(10)(9.80)(10)=980 J.PE_f=(10)(9.80)(10)=980\,\text{J}.

Conservation of mechanical energy then gives

KEf=1960−980=980 J.KE_f=1960-980=980\,\text{J}.

Therefore, after the rock has fallen 10 m:

  • Potential energy: 980 J980\,\text{J}
  • Kinetic energy: 980 J980\,\text{J}

Half of the initial gravitational potential energy has been converted into kinetic energy at this halfway height.

Evidence boundary

The original prompt itself states only the rock’s mass, the 20 m cliff, and the 10 m fall. The textbook’s worked solution then lists v1=0v_1=0 and g=9.80,mathrmm/s2g=9.80,mathrm{m/s^2}, and the surrounding section uses a frictionless approximation for these mechanical-energy transformations. The calculation follows that textbook model; with significant dissipative work, mechanical energy alone would not remain constant.

Sources

These references support the concepts and methods used in the explanation above.

Conservation of Energy Formula: Falling Rock | Verla