A frictionless plane is 15m long and inclined at 39.8 degrees. A sled starts at the bottom with an initial speed of 6.46m/s up the incline. When it reaches the point at which it momentarily stops, a second sled is released from the top of this incline with an initial speed vi. Both sleds reach the bottom of the incline at the same moment. The acceleration of gravity is 9.8m/s^2.
a) find the distance the the first sled traveled up the incline.
b) find the initial speed of the second sled
(a) The first sled stops momentarily when its kinetic energy is zero. At that time it has risen a vertical distance h such that
g h = Vo^2/2
h = Vo^2/(2g) = 2.13 m
distance up incline = h/sin39.8 = 3.33 m
(b) Compute the time it takes the first sled to go up (or down). Call that time T. Then compute how fast the second sled needs to be launched to get to the bottom in that time, traveling the full 15 m distance.
T*(Vo/2) = 3.33 m
T = 1.03 s
Take it from there
To solve this problem, we can use the principles of kinematics and energy conservation.
First, let's solve part (a) and find the distance the first sled traveled up the incline.
We can use the kinematic equation for displacement on an inclined plane:
s = ut + (1/2)at^2
Where:
s = displacement
u = initial velocity
a = acceleration
t = time
We know the initial velocity (u = 6.46 m/s) and the acceleration along the incline is due to gravity (a = 9.8 m/s^2). The sled eventually comes to a stop, so its final velocity (v) is 0 m/s. We also know that the sled reaches the bottom of the incline at the same time as the second sled, so the time taken by both sleds is the same.
Using the equation v = u + at and rearranging for time (t), we get:
t = (v - u) / a
Plugging in the values, we have:
t = (0 - 6.46) / (-9.8) ≈ 0.659 seconds
Now, let's find the distance traveled by the first sled using the formula for displacement:
s = ut + (1/2)at^2
Plugging in the values, we have:
s = 6.46 * 0.659 + (1/2) * (-9.8) * (0.659) ^ 2
Simplifying, we get:
s ≈ 2.131 meters
Therefore, the first sled traveled approximately 2.131 meters up the incline.
Now, let's move on and solve part (b) to find the initial speed of the second sled.
To conserve energy, the total mechanical energy at the top of the incline must be equal to the total mechanical energy at the bottom.
The mechanical energy at any point on the incline can be calculated as:
E = mgh + (1/2)mv^2
Where:
E = mechanical energy
m = mass of the sled
g = acceleration due to gravity
h = height above the reference point
v = velocity
For the first sled, the mechanical energy at the bottom is zero because it comes to a stop. The mechanical energy at the top can be calculated using the given information.
E_top1 = mgh_top1 + (1/2)mv_top1^2
Since the sled momentarily stops at the same point where the second sled is released, the height (h_top1) can be found using the inclined plane's height (h_plane) and angle A using trigonometry:
h_top1 = h_plane - (h_plane * sin(A))
Plugging in the values, we have:
h_top1 = 15 - (15 * sin(39.8)) ≈ 9.060 meters
The velocity (v_top1) at the top can be found using the kinematic equation v^2 = u^2 + 2as, where u is the initial velocity and s is the distance traveled up the incline:
v_top1^2 = u^2 + 2as
Plugging in the values, we have:
v_top1^2 = (6.46^2) + 2 * (-9.8) * 2.131
Simplifying, we get:
v_top1 ≈ 0.770 m/s
Now that we have the mechanical energy at the top (E_top1), which is equal to the mechanical energy at the bottom, we can calculate the initial speed of the second sled (v_i) using the following equation:
E_top1 = mgh_bottom + (1/2)mv_i^2
Since the height at the bottom (h_bottom) is zero, we have:
E_top1 = (1/2)mv_i^2
Plugging in the values, we have:
E_top1 = (1/2) * m * v_i^2
v_i^2 = (2 * E_top1) / m
Plugging in the values, we have:
v_i^2 = (2 * 0.770) / m
Since the initial speed of the second sled (v_i) is directly proportional to the square root of the expression, we can conclude:
v_i = √(2 * 0.770) ≈ 1.237 m/s
Therefore, the initial speed of the second sled is approximately 1.237 m/s.