QUESTION IMAGE
Question
- analyze a scenario where an objects potential energy decreases. what happens to its kinetic energy?
- how would the potential and kinetic energy of a skateboarder change as they go down a ramp?
Question 3
Step1: Recall Energy Conservation
In a closed system (ignoring non - conservative forces like friction), the total mechanical energy \(E = KE+PE\) is conserved, where \(KE\) is kinetic energy and \(PE\) is potential energy.
Step2: Analyze the Relationship
If the potential energy (\(PE\)) of an object decreases, according to the law of conservation of mechanical energy (\(E = KE + PE=\text{constant}\) in a closed system), the kinetic energy (\(KE\)) must increase. For example, when an object falls freely, its height (and thus gravitational potential energy \(PE = mgh\), where \(m\) is mass, \(g\) is acceleration due to gravity, \(h\) is height) decreases. As it falls, its speed increases, and since \(KE=\frac{1}{2}mv^{2}\) (where \(v\) is velocity), the kinetic energy increases.
Step1: Potential Energy Change
As the skateboarder goes down the ramp, the height \(h\) of the skateboarder above the ground (or a reference point) decreases. The gravitational potential energy is given by \(PE = mgh\). Since \(m\) (mass of skateboarder) and \(g\) (acceleration due to gravity) are constant, and \(h\) decreases, the potential energy (\(PE\)) of the skateboarder decreases.
Step2: Kinetic Energy Change
According to the law of conservation of mechanical energy (ignoring friction, which is a non - conservative force; in an ideal case), the total mechanical energy \(E=KE + PE\) is constant. Since \(PE\) is decreasing, the kinetic energy \(KE\) must increase. The kinetic energy is given by \(KE=\frac{1}{2}mv^{2}\), so the increase in kinetic energy is associated with an increase in the speed of the skateboarder as they go down the ramp.
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When an object's potential energy decreases (in a system with negligible non - conservative forces), its kinetic energy increases, as the total mechanical energy is conserved.