Questions: Now, consider the light boat runs ashore on a sand covered coral reef that has a 5 degree incline. Using energy considerations, calculate the distance the 8,000 kg light boat slides, given that its initial speed is 6.00 m / s and the force of friction acting between the ship and the sand is a constant 450 N. 24 m 20 m 22 m 17 m

Solution Steps

The problem involves a boat sliding up an incline, where we need to consider the initial kinetic energy, the work done against friction, and the gravitational potential energy gained. The initial kinetic energy (\( KE_i \)) is given by:

\[ KE_i = \frac{1}{2} m v^2 \]

where \( m = 8000 \, \text{kg} \) and \( v = 6.00 \, \text{m/s} \).

Substitute the given values into the kinetic energy formula:

\[ KE_i = \frac{1}{2} \times 8000 \, \text{kg} \times (6.00 \, \text{m/s})^2 = 144,000 \, \text{J} \]

The work done by friction (\( W_f \)) is given by:

\[ W_f = f \cdot d \]

where \( f = 450 \, \text{N} \) is the force of friction and \( d \) is the distance the boat slides. This work is negative because it opposes the motion.

The gravitational potential energy (\( PE \)) gained is:

\[ PE = mgh \]

where \( h = d \sin(\theta) \) and \( \theta = 5^\circ \).

The initial kinetic energy is converted into work done against friction and gravitational potential energy:

\[ KE_i = W_f + PE \]

Substitute the expressions for \( W_f \) and \( PE \):

\[ 144,000 = 450d + 8000 \times 9.81 \times d \sin(5^\circ) \]

First, calculate \( \sin(5^\circ) \):

\[ \sin(5^\circ) \approx 0.0872 \]

Substitute this into the equation:

\[ 144,000 = 450d + 8000 \times 9.81 \times 0.0872 \times d \]

Simplify:

\[ 144,000 = 450d + 6830.4d \]

\[ 144,000 = 7280.4d \]

Solve for \( d \):

\[ d = \frac{144,000}{7280.4} \approx 19.77 \, \text{m} \]

Final Answer