After performing a trick above the rim of a skateboard ramp, a 56 kg skateboarder lands on the ramp 3.5 m above ground level with a downward velocity of 4.0 m/s.
Ep = 0.0 J Ground level. Friction in the wheels of the skateboard and air resistance cause a loss of 9.0 × 10^2 J of mechanical energy. The skateboarder’s speed at the bottom of the ramp will be_____m/s.?
potential energy = m g h = 56*9.81 * 3.5
kinetic energy = (1/2) m v^2 = 28*16
sum to get total energy at top
kinetic energy at bottom = (1/2)m v^2
= total energy at top - 900
To find the skateboarder's speed at the bottom of the ramp, we need to use the principle of conservation of mechanical energy. The total mechanical energy, E, of the skateboarder is the sum of his potential energy, Ep, and his kinetic energy, Ek, at any given point.
E = Ep + Ek
At the top of the ramp, the skateboarder has only potential energy since he is not moving horizontally. The potential energy can be calculated using the formula:
Ep = m * g * h
where m is the mass of the skateboarder, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height above the ground.
Ep = 56 kg * 9.8 m/s^2 * 3.5 m
Ep = 1916.8 J
Since there is a loss of 900 J of mechanical energy due to friction and air resistance, we can calculate the final kinetic energy, Ek, at the bottom of the ramp by subtracting the loss from the initial mechanical energy:
Ek = E - loss
Ek = 1916.8 J - 900 J
Ek = 1016.8 J
The kinetic energy can also be expressed as:
Ek = 1/2 * m * v^2
where m is the mass of the skateboarder and v is his velocity.
To find the skateboarder's speed, we can rearrange the equation:
v = √(2 * Ek / m)
v = √(2 * 1016.8 J / 56 kg)
v ≈ √(36.3 m^2/s^2)
v ≈ 6.0 m/s
Therefore, the skateboarder's speed at the bottom of the ramp will be approximately 6.0 m/s.