A 4.0 kg crate slides from rest 6.0 m down a slope at 35°…

Step 1. Energy in. $h = 6.0\ \text{m} \sin 35^\circ = 3.44\ \text{m}$, so $mgh = 4.0\ \text{kg} \times 9.8\ \text{m/s}^2 \times 3.44\ \text{m} = 134.9\ \text{J}$.

Step 2. Heat. $0.22 \times 4.0\ \text{kg} \times 9.8\ \text{m/s}^2 \times \cos 35^\circ \times 6.0\ \text{m} = 42.4\ \text{J}$.

Step 3. Speed. $\tfrac12 \times 4.0\ \text{kg} \times v^2 = 134.9\ \text{J} - 42.4\ \text{J} = 92.5\ \text{J}$, so $v = 6.80\ \text{m/s}$.