Question

A group of bored kids at a skate park decide to build a launcher out of elastic exercise bands (i.e., stretchy strips of rubber or a bungee cords) and a wood ramp. they tie several bands to two closely spaced trees on a flat field and place the ramp in between the trees. the 50-kg "rider" stands on their skateboard at the bottom of the ramp as the other kids stretch the elastic bands back, hook them around the back of rider and then release the bands so that they snaps back, launching the rider up the ramp. approximate the bands as acting like a spring* with a spring constant of k = 1000 n/m that has been stretched by 2.0 meters. the ramp is angled at 30º above the horizontal and is 3.0 meters long. treat the skateboard rolling on the ramp as frictionless and neglect air resistance. calculate the horizontal distance between the back end of the ramp (the end where the rider goes airborne) and the location that the skateboard lands on the ground.

Answer 1

To calculate the horizontal distance between the back end of the ramp and the location that the skateboard lands on the ground, we can use the conservation of mechanical energy. Initially, all the potential energy is stored in the elastic bands, and when the rider is launched up the ramp, this potential energy is converted into kinetic energy.

Step-by-step explanation:

To calculate the horizontal distance between the back end of the ramp and the location that the skateboard lands on the ground, we can use the conservation of mechanical energy. Initially, all the potential energy is stored in the elastic bands, and when the rider is launched up the ramp, this potential energy is converted into kinetic energy.

The gravitational potential energy at the top of the ramp can be calculated using the formula PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. The height can be calculated as h = L*sin(theta), where L is the length of the ramp and theta is the angle. The kinetic energy at the top of the ramp can be calculated using the formula KE = (1/2)mv^2, where m is the mass and v is the velocity at the top of the ramp.

Since the ramp is frictionless, all the mechanical energy at the top of the ramp should be equal to the mechanical energy at the end of the ramp, which is only kinetic energy. Therefore, we can equate the expressions for kinetic energy at the top of the ramp and the kinetic energy at the end of the ramp to solve for the velocity, and then use this velocity to calculate the horizontal distance using the formula d = vt, where v is the velocity and t is the time.

Plugging in the given values, we get:

PE = mgh = (50 kg)(9.8 m/s^2)(3.0 m*sin(30°)) = 735 J

KE = (1/2)mv^2

Equating PE and KE:

mgh = (1/2)mv^2

735 J = (1/2)(50 kg)v^2

v = sqrt(2*735 J / 50 kg) ≈ 10.82 m/s

Now, we can use this velocity to calculate the horizontal distance:

d = vt = (10.82 m/s)(2.0 m) ≈ 21.64 m