Question

A 111 kg seal at an amusement park slides from rest down a ramp into the pool below. The top of the ramp is 1.90 m higher than the surface of the water and the ramp is inclined at an angle of 26.5° above the horizontal.

Answer 1

The final velocity of the seal just before entering the water is approximately 7.45 m/s.

Step-by-step explanation:

The problem involves the conservation of mechanical energy. Initially, the seal possesses gravitational potential energy, which is converted into kinetic energy as it slides down the ramp. The gravitational potential energy (GPE) at the top is given by the formula GPE = mgh, where m is the mass of the seal (111 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the height (1.90 m). The kinetic energy (KE) just before entering the water is given by
`\(KE = (1)/(2)mv^2\)`, where v is the final velocity.

At the top of the ramp, all the initial potential energy is converted into kinetic energy, so
`\(mgh = (1)/(2)mv^2\)`. Solving for v, we get

`\(v = √(2gh)\)`.

Substituting the values,

`\(v = √(2 * 9.8 * 1.90) \approx 7.45\) m/s`.

In this case, the angle of the ramp is not needed because the conservation of energy approach allows us to directly find the final velocity without considering the details of the path. The result is independent of the angle of the incline. The seal's final velocity just before entering the water is approximately 7.45 m/s.

Answer 2

The final speed of the seal as it reaches the surface of the water is approximately
`\(6.09 \, \text{m/s}\).`

Given:

- Mass of the seal (m) = 111 kg

- Change in height
`(\(\Delta h\))` = 1.90 m

- Acceleration due to gravity (\(g\)) = 9.81 m/s²

Calculate the Change in Potential Energy

`\[ \Delta PE = m \cdot g \cdot \Delta h \]`

`\[ \Delta PE = 111 \, \text{kg} * 9.81 \, \text{m/s}^2 * 1.90 \, \text{m} \]`

`\[ \Delta PE \approx 2061.75 \, \text{J} \]`

Convert the Change in Potential Energy to Kinetic Energy

Since energy is conserved, the change in potential energy is converted entirely into kinetic energy at the bottom of the ramp.

`\[ \Delta KE = \Delta PE = 2061.75 \, \text{J} \]`

Calculate the Final Speed Using Kinetic Energy

The kinetic energy at the bottom of the ramp
`(\( KE_{\text{bottom}} \))` is given by:

`\[ KE_{\text{bottom}} = (1)/(2) \cdot m \cdot v^2 \]`

Equating kinetic energy to the change in potential energy:

`\[ (1)/(2) \cdot m \cdot v^2 = \Delta PE \]`

`\[ (1)/(2) \cdot 111 \, \text{kg} \cdot v^2 = 2061.75 \, \text{J} \]`

`\[ v^2 = \frac{2 \cdot 2061.75 \, \text{J}}{111 \, \text{kg}} \]`

`\[ v^2 \approx 37.18 \, \text{m}^2/\text{s}^2 \]`

Taking the square root to find v:

`\[ v \approx √(37.18) \, \text{m/s} \]`

`\[ v \approx 6.09 \, \text{m/s} \]`

Therefore, after performing the full calculation, the final speed of the seal as it reaches the surface of the water is approximately
`\(6.09 \, \text{m/s}\).`

Question:

1) A 119 kg seal at an amusement park slides from rest down a ramp into the pool below. The top of the ramp is 1.80 m higher than the surface of the water and the ramp is inclined at an angle of 26.5 ∘ above the horizontal.

If the seal reaches the water with a speed of 4.55 m/s, what is the work done by kinetic friction?

w= _______ J

What is the coefficient of kinetic friction between the seal and the ramp?

u=________

2) A child swings back and forth on a swing suspended by 2.4 m -long ropes.

Find the turning-point angles if the child has a speed of 0.70 m/s when the ropes are vertical.

Theta +/-= ________ degrees