Question

using conservation of energy, find the speed $v_b$ of the block at the bottom of the ramp. express your answer in terms of some or all the variables $m$, $v$, and $h$ and any appropriate constants. view available hint(s) $v_b = $ submit

Explanation:

Step1: Identify Initial and Final Energies

Assume the block has an initial speed \( v \) (so initial kinetic energy \( KE_i = \frac{1}{2}mv^2 \)) and initial gravitational potential energy \( PE_i = mgh \) (taking the bottom of the ramp as the zero - potential level). At the bottom of the ramp, the potential energy \( PE_f = 0 \) and the kinetic energy \( KE_f=\frac{1}{2}mv_b^2 \). By the conservation of energy, the total initial energy equals the total final energy, so \( KE_i + PE_i=KE_f + PE_f \).

Step2: Substitute Energy Expressions

Substitute the energy formulas into the conservation of energy equation: \( \frac{1}{2}mv^2+mgh=\frac{1}{2}mv_b^2 + 0 \).

Step3: Solve for \( v_b \)

First, multiply both sides of the equation by 2 to get rid of the fractions: \( mv^2 + 2mgh=mv_b^2 \). Then, divide both sides by \( m \): \( v^2+2gh = v_b^2 \). Take the square root of both sides (we take the positive root since speed is a non - negative quantity): \( v_b=\sqrt{v^2 + 2gh} \).

Answer:

\(\sqrt{v^{2}+2gh}\)