A frictionless plane is 13.1 m long and inclined at 39.8 deg;. A sled starts at the bottom with an initial speed of 4.99 m/s up the incline. When the sled reaches the point at which it momentarily stops, a second sled is released from the top of the 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/s. find the distance the first sled traveled up the incline.

Question

A frictionless plane is 13.1 m long and inclined at 39.8 deg;. A sled starts at the bottom with an initial speed of 4.99 m/s up the incline. When the sled reaches the point at which it momentarily stops, a second sled is released from the top of the 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/s. find the distance the first sled traveled up the incline.

Answer 1

This problem has nothing to do with the second sled or the length of the plane as long as it does not overshoot and fly off the top.

Kinetic energy at bottom = potential energy stopped at top of trajectory.

(1/2) m Vi^2 = m g h = m g (x sin theta)

x = Vi^2 / (2 g sin theta)

Answer 2

To find the distance the first sled traveled up the incline, we need to apply the principles of motion along an inclined plane.

First, let's consider the motion of the sled that starts at the bottom with an initial speed of 4.99 m/s up the incline. We'll use the following variables:
- u1 = initial speed of the first sled = 4.99 m/s
- a = acceleration along the incline (which we need to find)
- θ = angle of inclination = 39.8 degrees
- x1 = distance traveled by the first sled up the incline (which we need to find)

Using the equation of motion along an inclined plane:
v^2 = u^2 + 2ax

We know that the sled momentarily stops at some point, so the final speed (v) would be zero for the first sled. Rearranging the equation, we have:
0 = (4.99 m/s)^2 + 2a(x1)

Simplifying this equation, we get:
2ax1 = -(4.99 m/s)^2

Now, let's consider the second sled that is released from the top of the incline with an initial speed vi. The distance traveled by the second sled down the incline is the length of the plane, which is given as 13.1 m.

Using the equation of motion along an inclined plane for the second sled:
v^2 = u^2 + 2ax

Once again, at the bottom of the incline, the final speed (v) would be zero for the second sled. Rearranging the equation, we have:
0 = (vi)^2 + 2a(13.1 m)

Simplifying this equation, we get:
2a(13.1 m) = - (vi)^2

Now, since both sleds reach the bottom of the incline at the same moment, we can equate the two expressions for a:
2ax1 = 2a(13.1 m)
-(4.99 m/s)^2 = (vi)^2

Applying the above equations, we can solve for x1, which is the distance traveled by the first sled up the incline.