Box Pulled Forces 84072D

1. **State the problem:** A $35kg$ box is pulled with a $250N$ tension force at an angle of $23^\circ$ above the horizontal. The coefficient of kinetic friction $\mu_k$ is $0.25$. We need to find the normal force, friction force, net force, acceleration, and describe the box's speed. 2. **Identify forces acting on the box:** There are 4 forces: - Tension force $T = 250N$ at $23^\circ$ - Weight $W = mg = 35 \times 9.8 = 343N$ downward - Normal force $N$ upward from the floor - Friction force $f_k$ opposing motion 3. **Calculate the normal force $N$:** The vertical forces balance: $$ N + T \sin 23^\circ = W $$ Rearranged: $$ N = W - T \sin 23^\circ $$ Calculate: $$ T \sin 23^\circ = 250 \times \sin 23^\circ \approx 250 \times 0.3907 = 97.68N $$ So: $$ N = 343 - 97.68 = 245.32N $$ 4. **Calculate the friction force $f_k$:** $$ f_k = \mu_k N = 0.25 \times 245.32 = 61.33N $$ 5. **Calculate the net force $F_{net}$:** Horizontal forces: $$ F_{net} = T \cos 23^\circ - f_k $$ Calculate: $$ T \cos 23^\circ = 250 \times \cos 23^\circ \approx 250 \times 0.9205 = 230.13N $$ So: $$ F_{net} = 230.13 - 61.33 = 168.8N $$ 6. **Calculate acceleration $a$:** Using Newton's second law: $$ a = \frac{F_{net}}{m} = \frac{168.8}{35} = 4.82 \text{ m/s}^2 $$ 7. **Describe the box's speed:** Since there is a net force forward, the box is accelerating and its speed is increasing in the direction of the pull. **Final answers:** - Number of forces: 4 - Normal force: $245.32N$ - Friction force: $61.33N$ - Net force: $168.8N$ - Acceleration: $4.82 \text{ m/s}^2$ - Speed: Increasing (accelerating) in the direction of the pull.