Question

work, energy, and power

1. a dart is launched from a dart gun and subsequently follows a parabolic path typical of any projectile. five positions in the trajectory of the dart are marked and labeled in the diagram at the right. for each of the five positions and for position z, fill in the work - energy bar chart in the space below.

z: position of dart when springs are compressed.
a: position of dart after release from springs.

1. a 2 - kg ball moving at 2 m/s is rolling towards an inclined plane. it eventually rolls up the hill to a position near the top where it momentarily stops prior to rolling back down the incline. assume negligible friction and air resistance. construct a energy bar chart for the ball.

simplify the equation below by canceling terms that are either zero or constant. then use the equation to determine the height to which the ball rises along the incline before stopping.
\\(\frac{1}{2} \cdot m \cdot v_{i}^{2} + m \cdot g \cdot h_{i} + f \cdot d \cdot \cos \theta = \frac{1}{2} \cdot m \cdot v_{f}^{2} + m \cdot g \cdot h_{f}\\)

1. three identical balls approach three different \frictionless\ hills with a speed of 2 m/s. in which case - a, b, or c, (or a tie) - will the ball roll the highest? ______ explain your answer.

© the physics classroom, 2009

Explanation:

Problem 4 (Simplify Equation and Find Height)

Step 1: Identify Zero/Constant Terms

• Friction is negligible, so \( F = 0 \) (force of friction), thus \( F \cdot d \cdot \cos\theta = 0 \).
• Final velocity \( v_f = 0 \) (ball stops), so \( \frac{1}{2} \cdot m \cdot v_f^2 = 0 \).
• Let initial height \( h_i = 0 \) (reference point), so \( m \cdot g \cdot h_i = 0 \).

The equation simplifies to:
\( \frac{1}{2} \cdot m \cdot v_i^2 = m \cdot g \cdot h_f \)

Step 2: Solve for \( h_f \)

Divide both sides by \( m \):
\( \frac{1}{2} \cdot v_i^2 = g \cdot h_f \)
Rearrange for \( h_f \):
\( h_f = \frac{v_i^2}{2g} \)

Substitute \( v_i = 2 \, \text{m/s} \), \( g = 9.8 \, \text{m/s}^2 \):
\( h_f = \frac{(2)^2}{2 \cdot 9.8} = \frac{4}{19.6} \approx 0.204 \, \text{m} \) (or \( \approx 0.2 \, \text{m} \))

Problem 5 (Which Ball Rolls Highest)

All balls have the same initial speed (\( 2 \, \text{m/s} \)) and mass (identical). With frictionless hills, mechanical energy (\( KE + PE \)) is conserved. Initial \( KE = \frac{1}{2}mv^2 \) is the same for all, and initial \( PE \) (if reference is ground) is same (or zero). At maximum height, \( KE = 0 \), so \( PE = mgh \) depends only on initial \( KE \), not hill shape. Thus, all balls rise to the same height (tie).

Answer:

(Problem 4):
Height \( \boldsymbol{\approx 0.20 \, \text{m}} \) (or \( \frac{2}{9.8} \approx 0.204 \, \text{m} \))