What is the final velocity of a cart at rest on a 94.68 m tall hill, assuming 6% of the energy is dissipated by friction?

To determine the final velocity of the cart at the bottom of the hill, we need to consider the conservation of energy, taking into account the energy lost due to friction.

First, let’s calculate the potential energy (PE) of the cart at the top of the hill. The potential energy is given by the formula:

PE = mgh

where:

  • m is the mass of the cart,
  • g is the acceleration due to gravity (approximately 9.81 m/s²),
  • h is the height of the hill (94.68 m).

Since the cart is at rest at the top of the hill, all its energy is in the form of potential energy. As the cart moves down the hill, this potential energy is converted into kinetic energy (KE), but some of it is lost due to friction.

Given that 6% of the energy is dissipated by friction, only 94% of the initial potential energy is converted into kinetic energy at the bottom of the hill. The kinetic energy is given by:

KE = 0.94 × PE

The kinetic energy can also be expressed in terms of the final velocity (v) of the cart:

KE = ½mv²

Setting the two expressions for KE equal to each other:

0.94 × mgh = ½mv²

We can cancel the mass (m) from both sides of the equation:

0.94 × gh = ½v²

Solving for the final velocity (v):

v² = 2 × 0.94 × gh

v = √(2 × 0.94 × gh)

Substituting the values for g and h:

v = √(2 × 0.94 × 9.81 × 94.68)

v ≈ √(2 × 0.94 × 9.81 × 94.68)

v ≈ √(1745.5)

v ≈ 41.78 m/s

Therefore, the final velocity of the cart at the bottom of the hill, assuming 6% of the energy is dissipated by friction, is approximately 41.78 m/s.

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