Question

haynes (hlh2749) - energy 1 - neff - (76523)
the acceleration of gravity is 9.81 m/s².
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024 (part 1 of 3) 10.0 points
the figure is a graph of the gravitational potential energy and kinetic energy of a 70 g yo-yo as it moves up and down on its string.
...... potential energy
—— kinetic energy

• - - mechanical energy

a) by what amount does the mechanical energy of the yo-yo change after 4.5 s?
answer in units of j.
025 (part 2 of 3) 10.0 points
b) what is the speed of the yo-yo after 7.5 s?
answer in units of m/s.
026 (part 3 of 3) 10.0 points
c) what is the maximum height of the yo-yo?
answer in units of m.
027 10.0 points
the sketch shows the potential energy u(r) between two particles. the total energy of the system of particles is denoted as e.

at which distance r between the particles do they have their maximum total kinetic energy?

1. r₂
2. only as r → ∞
3. near r = 0

Explanation:

Part a)

Step1: Recall Mechanical Energy Definition

Mechanical energy (\( E_{mech} \)) is the sum of kinetic energy (\( K \)) and potential energy (\( U \)), \( E_{mech} = K + U \). In a system with no non - conservative forces (or if non - conservative forces do no work), mechanical energy is conserved. From the graph, the mechanical energy is represented by the dashed line.

Step2: Analyze the Graph of Mechanical Energy

Looking at the graph of mechanical energy over time, we can see that the dashed line (mechanical energy) is horizontal. This means that the value of mechanical energy does not change with time. So, after \( 4.5 \) s, the change in mechanical energy \( \Delta E_{mech}=E_{final}-E_{initial} = 0 \), since \( E_{final}=E_{initial} \).

Part b)

Step1: Determine Mass and Mechanical Energy at \( t = 7.5 \) s

The mass of the yo - yo \( m = 70\space g=0.07\space kg \). From the graph, the mechanical energy at \( t = 7.5 \) s is \( E_{mech}=600\space mJ = 0.6\space J \) (since \( 1\space mJ = 10^{- 3}\space J \)). At the maximum height (when the yo - yo stops moving up or down), the kinetic energy \( K = 0 \), and mechanical energy is equal to potential energy. But when we want to find the speed at \( t = 7.5 \) s, we use the fact that at any time, \( E_{mech}=K + U \). However, when the yo - yo is at its lowest point (or when we consider the energy conservation), at \( t = 7.5 \) s, we can assume that the potential energy is at a minimum (or we can use the fact that mechanical energy is conserved and at some point we can relate it to kinetic energy). Wait, actually, from the graph, at \( t = 7.5 \) s, the mechanical energy is still \( 0.6\space J \). At the point when we want to find the speed, if we consider that at the lowest point (where potential energy is minimum, maybe zero? Wait, no, let's re - think. The mass is \( m = 0.07\space kg \), and \( E_{mech}=K=\frac{1}{2}mv^{2} \) (assuming that at the point where we calculate the speed, the potential energy is zero, or we can use the fact that mechanical energy is conserved and at \( t = 7.5 \) s, the mechanical energy is all kinetic? Wait, no, the graph shows kinetic and potential energy. Wait, the mechanical energy is constant (dashed line) at \( 600\space mJ = 0.6\space J \). So, \( E_{mech}=K=\frac{1}{2}mv^{2} \) (because when the yo - yo is moving, at some point, we can take the potential energy as zero for calculation, or maybe at the lowest point, potential energy is zero and kinetic energy is maximum, but here we can use \( E_{mech}=\frac{1}{2}mv^{2} \) to find \( v \).

Step2: Solve for Speed \( v \)

We know that \( E_{mech}=\frac{1}{2}mv^{2} \), so \( v=\sqrt{\frac{2E_{mech}}{m}} \). Substituting \( E_{mech} = 0.6\space J \) and \( m = 0.07\space kg \), we get \( v=\sqrt{\frac{2\times0.6}{0.07}}=\sqrt{\frac{1.2}{0.07}}\approx\sqrt{17.14}\approx 4.14 \)? Wait, no, maybe I made a mistake. Wait, the mass is \( 70\space g = 0.07\space kg \), and the mechanical energy is \( 600\space mJ=0.6\space J \). Wait, let's check the graph again. The kinetic energy graph: at \( t = 7.5 \) s, the kinetic energy is about \( 400\space mJ = 0.4\space J \)? No, the mechanical energy is the dashed line, which is at \( 600\space mJ \). Wait, maybe the yo - yo's mass is \( 70\space g = 0.07\space kg \), and we use \( E_{mech}=mgh+\frac{1}{2}mv^{2} \), but at the maximum height, \( v = 0 \), so \( E_{mech}=mgh_{max} \). But when we want to find the speed at \( t = 7.5 \) s, we can use \( E_{mech}=\frac{1}{2}mv^{2}+mgh \). But maybe a better approach: from the graph, at \( t = 7.5 \) s, the kinetic energy can be found? Wait, no, the mechanical energy is constant. Wait, let's recalculate: \( E_{mech}=0.6\space J \), \( m = 0.07\space kg \). If we assume that at the point where we calculate the speed, the potential energy is zero (which might be the case at the lowest point), then \( E_{mech}=\frac{1}{2}mv^{2} \), so \( v=\sqrt{\frac{2E_{mech}}{m}}=\sqrt{\frac{2\times0.6}{0.07}}=\sqrt{\frac{1.2}{0.07}}\approx\sqrt{17.14}\approx 4.14 \) m/s. But maybe I misread the graph. Wait, the mass is \( 70\space g = 0.07\space kg \), and the mechanical energy is \( 600\space mJ = 0.6\space J \). Wait, another way: at \( t = 0 \), the yo - yo starts from rest, so \( K = 0 \), \( E_{mech}=U = 600\space mJ \). Then, when it moves, the mechanical energ…

Step1: Recall Energy - Height Relationship

The gravitational potential energy is given by \( U = mgh \), and at the maximum height \( h_{max} \), the kinetic energy \( K = 0 \), so the mechanical energy \( E_{mech}=U = mgh_{max} \).

Step2: Solve for \( h_{max} \)

We know that \( E_{mech}=600\space mJ = 0.6\space J \), \( m = 0.07\space kg \), and \( g = 9.81\space m/s^{2} \). Rearranging the formula \( h_{max}=\frac{E_{mech}}{mg} \). Substituting the values: \( h_{max}=\frac{0.6}{0.07\times9.81}=\frac{0.6}{0.6867}\approx0.87\space m \).

Part 027

Answer:

\( 0 \) J