The 20lb cartis released from rest at time t=0 on the incline surface. The force P=8t lb acts on the cart, where t is the time measured in seconds. A) Determine the distance the cart will move down the inclined surface before reversing direction. B) Find the velocity of the cart when it returns to the point of release.
The angle of inclination is 20 degrees
I assume that the force is acting up along the plane.
Since F = ma, a = 0.4t - 9.8sin20° = 0.4t - 3.35 ft/s^2
v(t) = 0.2t^2 - 3.35t
so, v=0 when t=16.76 s
s(t) = 1/15 t^3 - 1.675t^2
Now just figure how far it has gone (s(16.76)), and how long it will take to get back up the plane.
To solve this problem, we need to determine the distance the cart will move down the inclined surface before reversing direction. We can find the distance by integrating the acceleration with respect to time, and then solving for the time when the velocity reaches zero.
First, let's find the acceleration of the cart. The force acting on the cart is P = 8t lb, and the mass of the cart is 20 lb. We can use Newton's second law, F = ma, to find the acceleration:
8t = 20 * a
Solving for a, we get:
a = (8t) / 20
Next, let's integrate the acceleration with respect to time to find the velocity:
v = ∫a dt
Integrating (8t) / 20 with respect to t, we get:
v = (4t^2) / 20 + C
where C is the constant of integration.
Now, let's find the time when the velocity reaches zero. We can set the velocity equation equal to zero and solve for t:
0 = (4t^2) / 20 + C
(4t^2) / 20 = -C
4t^2 = -20C
t^2 = -5C
Since C is a constant, we know that (-5C) is also a constant. Let's call it K:
t^2 = K
Taking the square root of both sides, we get:
t = √K
Now, let's determine the value of K. The velocity is zero when the cart returns to the point of release, which means t = 0. Substituting t = 0 into the velocity equation, we get:
0 = (4*0^2) / 20 + C
0 = 0 + C
C = 0
Therefore, K = -5C = -5(0) = 0.
Since K is zero, the time when the velocity reaches zero is t = 0.
Now, let's determine the distance the cart will move down the inclined surface before reversing direction. We can find the distance by integrating the velocity with respect to time:
s = ∫v dt
Integrating (4t^2) / 20 with respect to t, we get:
s = (2t^3) / 20 + Ct + D
where C and D are constants of integration.
Since the cart is released from rest at t = 0, the initial velocity is zero and the cart starts from the point of release, so D = 0.
Now, let's find the position when the velocity is zero, which corresponds to the time t = 0:
s = (2*0^3) / 20 + C(0) + 0
s = 0 + 0 + 0
s = 0
Therefore, the distance the cart will move down the inclined surface before reversing direction is 0.
To summarize:
A) The distance the cart will move down the inclined surface before reversing direction is 0.
B) The velocity of the cart when it returns to the point of release is 0.
To solve this problem, we can break it down into two parts: finding the distance the cart will move down the inclined surface before reversing direction, and finding the velocity of the cart when it returns to the point of release.
A) To determine the distance the cart will move down the inclined surface before reversing direction, we need to integrate the force acting on the cart with respect to time, and then determine the displacement.
1. First, let's find the acceleration of the cart. We can use Newton's second law, which states that the net force on an object is equal to its mass times its acceleration (F = ma). In this case, the force acting on the cart is given by P = 8t lb.
Since the force acting on the cart is parallel to the incline, we can resolve it into two components: one parallel to the incline (Psinθ) and one perpendicular to the incline (Pcosθ), where θ is the angle of inclination.
The perpendicular component (Pcosθ) doesn't affect the motion of the cart along the incline, so we'll focus on the parallel component (Psinθ).
2. The net force acting on the cart along the incline is given by F = m * a, where m is the mass of the cart. In this case, the mass is 20 lb.
We can equate the force Psinθ to m * a and solve for a:
Psinθ = m * a
8t * sinθ = 20 * a
a = (8t * sinθ) / 20
3. To find the distance moved by the cart down the incline, we need to integrate the velocity with respect to time and then determine the displacement.
The velocity v of the cart can be found by integrating the acceleration a with respect to time:
v = ∫ a dt
Integrating the equation a = (8t * sinθ) / 20 with respect to t, we get:
v = ∫ (8t * sinθ) / 20 dt
v = (4 * sinθ / 20) * t^2
To find the distance moved down the incline, we need to determine the time at which the velocity becomes zero. This is the time when the cart reverses its direction.
Setting v = 0 and solving for t:
(4 * sinθ / 20) * t^2 = 0
t = 0 or t = √(20 / (4 * sinθ))
To find the distance d, we can integrate the velocity with respect to time from t = 0 to t = √(20 / (4 * sinθ)):
d = ∫ v dt
d = ∫ (4 * sinθ / 20) * t^2 dt
d = (4 * sinθ / 20) * (t^3 / 3) evaluated from 0 to √(20 / (4 * sinθ))
d = (4 * sinθ / 20) * (√(20 / (4 * sinθ))^3 / 3)
Simplifying the expression, we get the distance moved down the incline before reversing direction.
B) To find the velocity of the cart when it returns to the point of release, we can substitute the time at which it reverses its direction (t = √(20 / (4 * sinθ))) into the equation for the velocity:
v = (4 * sinθ / 20) * t^2
Solving for v, we can find the velocity of the cart when it returns to the point of release.