1. **State the problem:** A farmer pulls a sled loaded with firewood a distance of 20 m along level ground. The sled and load weigh 14,700 N. The tractor pulls with a 5000-N force at an angle above the horizontal, and friction opposes motion with a 3500-N force. We need to find the work done by each force and the total work done. 2. **Identify forces and work formula:** Work done by a force is given by $$W = Fd\cos(\theta)$$ where $F$ is the force magnitude, $d$ is displacement, and $\theta$ is the angle between force and displacement. 3. **Work done by the tractor:** The tractor force $F_t = 5000$ N at angle $\theta$ (angle not specified, assume $\theta$ degrees). Work by tractor: $$W_t = 5000 \times 20 \times \cos(\theta) = 100000 \cos(\theta)$$ J. 4. **Work done by friction:** Friction force $F_f = 3500$ N opposes motion, so angle between friction and displacement is 180°. Work by friction: $$W_f = 3500 \times 20 \times \cos(180^\circ) = 70000 \times (-1) = -70000$$ J. 5. **Work done by weight:** Weight acts vertically downward, displacement is horizontal, so angle is 90°. Work by weight: $$W_w = 14700 \times 20 \times \cos(90^\circ) = 0$$ J. 6. **Work done by normal force:** Normal force is vertical upward, displacement horizontal, angle 90°, so work is zero. 7. **Total work done:** Sum of all works: $$W_{total} = W_t + W_f + W_w + W_n = 100000 \cos(\theta) - 70000 + 0 + 0 = 100000 \cos(\theta) - 70000$$ J. **Note:** The problem does not specify the angle $\theta$ of the tractor force, so the answer is expressed in terms of $\cos(\theta)$. **Final answers:** - Work by tractor: $$100000 \cos(\theta)$$ J - Work by friction: $$-70000$$ J - Work by weight: $$0$$ J - Work by normal force: $$0$$ J - Total work: $$100000 \cos(\theta) - 70000$$ J