a skaterboarder is moving 5.25 m/s when he starts to roll up a frictionless hill. how much higher is he when he comes to a stop? (energy problem)
all the KE (1/2 mv^2) is converted to PE (mgh) So,
1/2 mv^2 = mgh
v^2 = 2gh
h = v^2/2g
To solve this problem, we can use the principle of conservation of mechanical energy, which states that the total mechanical energy of a system remains constant if no external forces are acting on it. The mechanical energy of a system consists of kinetic energy (KE) and potential energy (PE).
Let's break down the problem step by step:
1. We know that the skater starts with kinetic energy due to their motion. The formula for kinetic energy is KE = (1/2)mv^2, where m is the mass of the skater (which we assume to be constant) and v is the velocity.
2. The skater comes to a stop at the top of the hill, which means all of their kinetic energy is converted into potential energy. The formula for potential energy is PE = mgh, where m is the mass of the skater, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
3. Since mechanical energy is conserved, the initial kinetic energy (KE_initial) must be equal to the final potential energy (PE_final). Therefore, (1/2)mv^2 = mgh.
4. We can cancel out the mass (m) on both sides of the equation, giving us (1/2)v^2 = gh.
5. Now we can solve for h by rearranging the equation: h = (1/2)v^2/g.
6. Plug in the given values: v = 5.25 m/s and g = 9.8 m/s^2.
h = (1/2) * (5.25 m/s)^2 / (9.8 m/s^2)
= 13.734 m
Therefore, the skater is approximately 13.734 meters higher when he comes to a stop.