Problem
17)Group A measures the speed of a $0.1 \mathrm{~kg}$ car at $2.5 \mathrm{~m} / \mathrm{s}$ at the bottom of the hill. * Group B measures the speed of the same car at $3 \mathrm{~m} / \mathrm{s}$ at the bottom of the hill. The car started at the top of the hill $1 \mathrm{~m}$ high. What is the kinetic energy of the car for group A ? $0.63 \mathrm{~J}$ $0.31 J$ $0.45 \mathrm{~J}$
Solution
Step 1 ：Calculate the kinetic energy using the formula KE = 0.5 * m * v^2, where m is the mass of the car (0.1 kg) and v is its velocity (2.5 m/s): KE = 0.5 * 0.1 * (2.5)^2
Step 2 ：Simplify the expression to get the final kinetic energy: \(\boxed{0.31 J}\)
