1. **Stating the problem:** A 5 kg object is pulled upward with an acceleration of 0.3 m/s² by a rope. We need to find the tension in the rope and explain if the object feels lighter or heavier. 2. **Step 1: Write down forces and knowns:** Mass $m=5$ kg, acceleration upward $a=0.3$ m/s², gravitational acceleration $g=9.8$ m/s². 3. **Step 2: Calculate weight:** Weight force $F_w = mg = 5 \times 9.8 = 49$ N downward. 4. **Step 3: Apply Newton's 2nd law on vertical forces:** Let tension be $T$ upward. $$\sum F = ma \Rightarrow T - mg = ma$$ 5. **Step 4: Solve for tension:** $$T = mg + ma = m(g + a) = 5(9.8 + 0.3) = 5 \times 10.1 = 50.5 \text{ N}$$ 6. **Step 5: Interpretation:** The tension is greater than the weight, so the object feels heavier because the rope must not only balance gravity but also provide upward acceleration. --- 7. **Second problem:** An 8 kg box on a ramp inclined at $33^\circ$, coefficient of kinetic friction $_k = 0.3$, box is pushed up with acceleration 3.6 m/s² by a horizontal force $F$. Find $F$. 8. **Step 1: Define forces:** - Gravity $mg = 8 \times 9.8 = 78.4$ N downward. - Ramp angle $\theta = 33^\circ$. - Coefficient of kinetic friction $\mu_k = 0.3$. - Acceleration $a=3.6$ m/s² up the ramp. 9. **Step 2: Resolve weight into components:** Parallel to ramp (downhill): $$F_{g\parallel} = mg \sin(\theta) = 78.4 \times \sin(33^\circ) \approx 78.4 \times 0.5446 = 42.69 \text{ N}$$ Perpendicular to ramp: $$F_{g\perp} = mg \cos(\theta) = 78.4 \times \cos(33^\circ) \approx 78.4 \times 0.8387 = 65.7 \text{ N}$$ 10. **Step 3: Calculate friction force:** $$F_f = \mu_k \times F_{g\perp} = 0.3 \times 65.7 = 19.7 \text{ N}$$ 11. **Step 4: Find normal force $N$ taking horizontal force $F$ into account:** The box is pushed by force $F$ horizontally. Let's define axes along the ramp (parallel) and perpendicular to ramp. The force $F$ has components: - Along ramp upward: $F \cos(33^\circ)$ - Perpendicular to ramp inward or outward: $F \sin(33^\circ)$ (acting perpendicular to the ramp surface). Normal force is balanced by: $$N = F_{g\perp} - F \sin(33^\circ)$$ Friction depends on $N$: $$F_f = \mu_k N = \mu_k (F_{g\perp} - F \sin(33^\circ))$$ 12. **Step 5: Write equation along ramp for net force:** Acceleration up the ramp $a=3.6$ m/s², mass $m=8$ kg Net force $F_{net} = ma = m a = 8 \times 3.6 = 28.8$ N Sum of forces along ramp is applied horizontal force component minus weight component down ramp minus friction: $$F \cos(33^\circ) - F_{g\parallel} - F_f = ma$$ Insert friction from Step 11: $$F \cos(33^\circ) - 42.69 - 0.3 (65.7 - F \sin(33^\circ)) = 28.8$$ 13. **Step 6: Simplify and solve for $F$:** $$F \cos(33^\circ) - 42.69 - 0.3 \times 65.7 + 0.3 F \sin(33^\circ) = 28.8$$ Calculate constants: $$0.3 \times 65.7 = 19.7$$ Rewrite: $$F \cos(33^\circ) + 0.3 F \sin(33^\circ) = 28.8 + 42.69 + 19.7$$ Calculate right side sum: $$28.8 + 42.69 + 19.7 = 91.19$$ Calculate trig values: $$\cos(33^\circ) \approx 0.8387, \quad \sin(33^\circ) \approx 0.5446$$ Group terms: $$F (0.8387 + 0.3 \times 0.5446) = 91.19$$ Calculate inside parenthesis: $$0.8387 + 0.3 \times 0.5446 = 0.8387 + 0.1634 = 1.0021$$ So: $$F \times 1.0021 = 91.19 \Rightarrow F = \frac{91.19}{1.0021} \approx 91.0 \text{ N}$$ **Final answer:** $$\boxed{F \approx 91.0 \text{ N}}$$ --- 14. **Third problem:** Newton's 3rd law for astronaut drifting and Newton's 2nd law for acceleration. **Part (a):** To move back to the shuttle when tether broken, the best method is: iii. Take a tool from her tool belt and throw it away from the shuttle. **Explanation:** According to Newton's 3rd law, every action has an equal and opposite reaction. Throwing the tool away imparts momentum opposite to the shuttle direction, propelling the astronaut toward the shuttle. **Part (b):** Given astronaut mass $m$ and applied force $F$, acceleration is $$a = \frac{F}{m}$$ **Summary:** Using Newton's laws gives the best method for movement and the formula to calculate her acceleration based on force.