Gayle runs at a speed of 3.80 m/s and dives on a sled, initially at rest on the top of a frictionless snow covered hill. After she has descended a vertical distance of 5.00 m, her brother, who is initially at rest, hops on her back and together they continue down the hill. What is their speed at the bottom of the hill if the total vertical drop is 15.0 m? Gayle's mass is 50.0 kg, the sled has a mass of 5.00 kg and her brother has a mass of 30.0 kg.
To solve this problem, we need to apply the law of conservation of mechanical energy, which states that the total mechanical energy of a system remains constant if there are no external forces. The initial mechanical energy is equal to the final mechanical energy.
The initial mechanical energy is given by the sum of the potential energy and the kinetic energy.
The potential energy is mgh, where m is the total mass, g is the acceleration due to gravity, and h is the vertical distance.
The kinetic energy is given by the formula 0.5mv^2, where m is the total mass and v is the velocity.
Let's calculate the initial mechanical energy (Em_initial) and final mechanical energy (Em_final) for the given situation.
Em_initial = m_1 * g * h_1 + (m_2+m_3) * 0.5 * v_i^2
Em_final = m_1 * g * h_2 + (m_2+m_3) * 0.5 * v_f^2
where:
m_1 = 50.0 kg (Gayle's mass)
m_2 = 5.00 kg (mass of the sled)
m_3 = 30.0 kg (brother's mass)
v_i = 3.80 m/s (initial velocity)
h_1 = 5.00 m (initial vertical distance)
h_2 = 15.0 m (final vertical distance)
g = 9.8 m/s^2 (acceleration due to gravity)
Now, set the initial mechanical energy equal to the final mechanical energy and solve for the final velocity (v_f):
m_1 * g * h_1 + (m_2+m_3) * 0.5 * v_i^2 = m_1 * g * h_2 + (m_2+m_3) * 0.5 * v_f^2
Substituting the given values:
50.0 kg * 9.8 m/s^2 * 5.00 m + (5.00 kg + 30.0 kg) * 0.5 * (3.80 m/s)^2 = 50.0 kg * 9.8 m/s^2 * 15.0 m + (5.00 kg + 30.0 kg) * 0.5 * v_f^2
Simplify the equation and solve for v_f:
2450 J + 305 J = 7350 J + 35 v_f^2
2755 = 7350 + 35 v_f^2
35 v_f^2 = -4595
v_f^2 = - 131.286
v_f = sqrt(-131.286)
Since we obtained a negative value for the final velocity, it means that the expression inside the square root is not physically possible.
Therefore, there must be an error in the calculation or in the given values. Please double-check the given values and try solving the problem again.
To find the speed of Gayle and her brother at the bottom of the hill, we can use the principle of conservation of mechanical energy. The total mechanical energy at the top of the hill is equal to the total mechanical energy at the bottom of the hill.
At the top of the hill, Gayle's initial mechanical energy consists of both kinetic energy and potential energy. Since she is initially at rest, her initial kinetic energy is zero. Her initial potential energy is given by:
PEi = m * g * h
Where m is Gayle's mass, g is the acceleration due to gravity (9.8 m/s²), and h is the initial vertical height (5.00 m).
PEi = 50.0 kg * 9.8 m/s² * 5.00 m
PEi = 2450 J
At the bottom of the hill, the final mechanical energy consists only of kinetic energy. The total mass of Gayle, the sled, and her brother is 50.0 kg + 5.00 kg + 30.0 kg = 85.0 kg. The final vertical height is 15.0 m.
Using the principle of conservation of mechanical energy, we can equate the potential energy at the top of the hill to the kinetic energy at the bottom:
PEi = KEf
2450 J = KEf
We can now solve for the final kinetic energy:
KEf = 2450 J
Since kinetic energy is given by the equation:
KEf = (1/2) * m * v²
where m is the total mass (85.0 kg) and v is the final velocity, we can rearrange the equation to solve for v:
v = √(2 * KEf / m)
Substituting the values:
v = √(2 * 2450 J / 85.0 kg)
v ≈ √(57.65 m²/s²)
v ≈ 7.59 m/s
Therefore, the speed of Gayle and her brother at the bottom of the hill is approximately 7.59 m/s.
1. figure with conservation of momentum the initial velocity of gayle/sled
2. Now, with conservation of energy, the speed at bottome of hill.
KEbottom=KEattop+PEtop
1/2(Mg+ms)vbottom^2=1/2(Mg+ms)Vtop^2 + (Mg+Ms)g*height
figure vbottom.
Now, with conservation of momentum, figure the new velocity with brother on back.
Then, at the bottom of the hill, figure new velocity with conservation of energy, as before on the first hill.