Let's analyze the problem step by step: --- # Energy Bar Chart Construction ## Initial Energy - block of mass \( m \) starts at rest, so initial kinetic energy KE_{initial} = ). - The has potential energy due to its height \( d \): \[ PE_{initial} = m g d \] The spring is by a distance \( x \), storing elastic potential energy: \[ PE_{spring} = \frac{1}{2} k x^2 \] **Initial total energy:** \[ E_{initial} = PE_{initial} + PE_{spring} = m g d + \frac{1}{2} k x^2 \] --- ## Final Energy - After the block slides a distance \( d \), it has velocity \( v_f \). - The potential energy due to height becomes zero (assuming it slides onto the surface at ground level). - The spring is no longer compressed, so elastic potential energy is zero. - The kinetic energy of the block is: \[ KE_{final} = \frac{1}{2} m v_f^2 \] - The surface has a coefficient of kinetic friction \( \mu_k \), which dissipates energy as work done against friction: \[ W_{friction} = \mu_k m g d \] --- # Energy Bar Chart ### Initial state: - Elastic potential energy: \( \frac{1}{2} k x^2 \) - Gravitational potential energy: \( m g d \) - Kinetic energy: ### Final state: - Kinetic energy: \( \frac{1}{2} m v_f^2 \) - Work done against friction: \( \mu_k m g d \) --- # Energy Conservation Equation Using conservation of energy, the initial energy minus the work done by friction equals the final kinetic energy: \[ m g d + \frac{1}{2} k x^2 = \frac{1}{2} m v_f^2 + \mu_k m g d \] --- # Summary The key equation describing the situation is: \[ \boxed{ \frac{1}{2} m v_f^2 = m g d + \frac{1}{2} k x^2 - \mu_k m g d } \] or equivalently: \[ \boxed{ \frac{1}{2} m v_f^2 = (1 - \mu_k) m g d + \frac{1}{2} k x^2 } \] This equation accounts for the initial energy, energy lost due to friction, and the kinetic energy after sliding. --- Feel free to ask if you'd like assistance with drawing the energy bar chart!