Total Force 9Fd77D

1. **State the problem:** We have a box of mass $60$ kg pushed up a ramp inclined at $30^\circ$ with a pushing force of $400$ N and a friction coefficient $k=0.07$. We want to find the total force acting on the box. 2. **Identify forces involved:** - Gravitational force component down the ramp: $F_g = mg \sin(\theta)$ - Normal force: $N = mg \cos(\theta)$ - Frictional force: $F_f = kN = k mg \cos(\theta)$ - Applied force: $F_a = 400$ N (up the ramp) 3. **Calculate each force:** - Weight $mg = 60 \times 9.8 = 588$ N - Gravitational component down the ramp: $$F_g = 588 \times 0.5 = 294 \text{ N}$$ - Normal force: $$N = 588 \times 0.866 = 509.208 \text{ N}$$ - Frictional force: $$F_f = 0.07 \times 509.208 = 35.64456 \text{ N}$$ 4. **Calculate total force:** The total force $F_{total}$ is the applied force minus the forces opposing motion (gravity component and friction): $$F_{total} = F_a - F_g - F_f = 400 - 294 - 35.64456 = 70.35544 \text{ N}$$ 5. **Round the answer:** $$F_{total} \approx 70.36 \text{ N}$$ 6. **Answer choice:** The closest answer is C. 70.36 N. **Final answer:** $\boxed{70.36 \text{ N}}$