During a rockslide, a 740 kg rock slides from rest down a hillside that is 500 m long and 300 m high. The coefficient of kinetic friction between the rock and the hill surface is 0.18.
(a) If the gravitational potential energy U of the rock-Earth system is set to zero at the bottom of the hill, what is the value of U just before the slide? (b) How much energy is transferred to thermal energy during the slide? (c) What is the kinetic energy of the rock as it reaches the bottom of the hill?(d) What is its speed then?
To answer these questions, we can use the principles of work and energy.
(a) The gravitational potential energy (U) of the rock-Earth system just before the slide can be calculated using the formula U = mgh, where m is the mass of the rock, g is the acceleration due to gravity, and h is the height of the hill. Given that the mass of the rock (m) is 740 kg and the height of the hill (h) is 300 m, we have:
U = (740 kg) * (9.8 m/s^2) * (300 m)
U = 2,165,200 J
Therefore, the value of U just before the slide is 2,165,200 Joules.
(b) The energy transferred to thermal energy during the slide can be calculated using the work-energy principle. The work done by the friction force can be found using the formula W_friction = μ * m * g * d, where μ is the coefficient of kinetic friction, m is the mass of the rock, g is the acceleration due to gravity, and d is the distance of the slide. Given that the coefficient of kinetic friction (μ) is 0.18, the mass of the rock (m) is 740 kg, the acceleration due to gravity (g) is 9.8 m/s^2, and the distance of the slide (d) is 500 m, we have:
W_friction = (0.18) * (740 kg) * (9.8 m/s^2) * (500 m)
W_friction = 646,800 J
Therefore, the energy transferred to thermal energy during the slide is 646,800 Joules.
(c) The kinetic energy (K) of the rock as it reaches the bottom of the hill can be calculated using the principle of conservation of energy. Since the initial potential energy (U_i) is converted into the sum of final kinetic energy (K_f) and final thermal energy (Q), we can write:
U_i = K_f + Q
Since the potential energy just before the slide (U_i) was calculated to be 2,165,200 Joules and the energy transferred to thermal energy during the slide (Q) was calculated to be 646,800 Joules, we have:
K_f = U_i - Q
K_f = 2,165,200 J - 646,800 J
K_f = 1,518,400 J
Therefore, the kinetic energy of the rock as it reaches the bottom of the hill is 1,518,400 Joules.
(d) The speed (v) of the rock at the bottom of the hill can be calculated using the kinetic energy. The formula for kinetic energy can be rearranged as:
K = 0.5 * m * v^2
Plugging in the values of the kinetic energy (K) as calculated in part(c) and the mass of the rock (m) as given (740 kg), we have:
1,518,400 J = 0.5 * (740 kg) * v^2
Simplifying the equation, we can solve for v:
v^2 = (2 * 1,518,400 J) / (740 kg)
v^2 = 4,097.297
Taking the square root of both sides, we find:
v ≈ 64.0 m/s
Therefore, the speed of the rock at the bottom of the hill is approximately 64.0 m/s.
To answer these questions, we need to calculate the gravitational potential energy (U), the energy transferred to thermal energy, the kinetic energy, and the speed of the rock at the bottom of the hill. Let's go step by step:
(a) To find the value of gravitational potential energy (U) just before the slide, we can use the formula:
U = mgh,
where m is the mass of the rock, g is the acceleration due to gravity, and h is the height.
Given:
m = 740 kg,
g = 9.8 m/s^2,
h = 300 m.
Plugging in the values into the formula:
U = 740 kg * 9.8 m/s^2 * 300 m = 2,166,000 Joules.
Therefore, the gravitational potential energy just before the slide is 2,166,000 Joules.
(b) The energy transferred to thermal energy during the slide can be determined by the work-energy principle. The total work done on the rock is equal to the change in its mechanical energy (potential energy + kinetic energy). In this case, since the rock starts from rest, its initial kinetic energy is zero.
The work done by the force of friction is given by:
W = μk * m * g * d,
where μk is the coefficient of kinetic friction, m is the mass of the rock, g is the acceleration due to gravity, and d is the distance.
Given:
μk = 0.18,
m = 740 kg,
g = 9.8 m/s^2,
d = 500 m.
Plugging in the values into the formula:
W = 0.18 * 740 kg * 9.8 m/s^2 * 500 m = 647,400 Joules.
Therefore, the energy transferred to thermal energy during the slide is 647,400 Joules.
(c) The kinetic energy (K) of the rock as it reaches the bottom of the hill can be calculated using the formula:
K = (1/2) * m * v^2,
where m is the mass of the rock and v is its velocity.
Given:
m = 740 kg.
We still need to find v, the final velocity of the rock at the bottom of the hill.
(d) To find the speed of the rock at the bottom of the hill, we can use the principle of conservation of energy. The mechanical energy at the top of the hill (potential energy) is equal to the mechanical energy at the bottom of the hill (kinetic energy).
Using the formula:
U_top + K_top = U_bottom + K_bottom,
where U is the gravitational potential energy and K is the kinetic energy.
Given:
U_top = 2,166,000 Joules (from part a),
K_top = 0 Joules (since the rock starts from rest),
U_bottom = 0 Joules (reference point),
K_bottom = ?
The equation becomes:
0 + 0 = 0 + K_bottom.
Therefore, K_bottom = 0 Joules.
Since the kinetic energy at the bottom of the hill is zero, the speed (v) is also zero.
In summary:
(a) The value of U just before the slide is 2,166,000 Joules.
(b) The energy transferred to thermal energy during the slide is 647,400 Joules.
(c) The kinetic energy of the rock as it reaches the bottom of the hill is 0 Joules.
(d) The speed of the rock at the bottom of the hill is 0 m/s.