High School

A 58-kg rock climber at rest loses her control and starts to slide down through her rope from 186 m above the land shelf. She lands to the shelf with a velocity of 23m/s. Find the work done by the friction until she lands the shelf.

Answer :

The work done by friction until the rock climber lands on the shelf is approximately 105468.8 Joules.

To find the work done by friction on the rock climber, we need to calculate the change in the gravitational potential energy of the climber as she slides down.

The change in gravitational potential energy is given by the formula:

ΔPE = m * g * Δh

where:

ΔPE is the change in gravitational potential energy,

m is the mass of the rock climber (58 kg),

g is the acceleration due to gravity (approximately 9.8 m/s²), and

Δh is the change in height (186 m).

Substituting the values into the formula, we have:

ΔPE = 58 kg * 9.8 m/s² * (-186 m)

The negative sign indicates that the gravitational potential energy decreases as the climber descends.

Calculating the value, we find:

ΔPE = -105468.8 J

The work done by friction is equal to the change in gravitational potential energy, but with a positive sign since friction acts in the direction of the displacement. Therefore, the work done by friction is:

Work = |ΔPE| = 105468.8 J

To know more about gravitational potential energy

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