Show your work-formulas,calculations, and diagrams-for the remaining questions. me same as that duiling on the builet. 3. Om and Rovic are helping Dilshan push his car out of a snow drift where there is some friction due to the sheer size of the pile of snow. The 1555 kg car starts to accelerate from rest during 3.00 seconds of solid pushing. a. Draw a free -body diagram.Label all forces and their relation to each other. b. What was the acceleration of the car if the net force was 25 N?/1 c. The coefficient of friction on the snow-covered road was found to be 0.320. What force must the three have been applying to achieve this net force and acceleration? d. How far could the car go, from rest, during this push?12

**3a. Free-Body Diagram:**<br /><br />[asy]<br />unitsize(0.5 cm);<br /><br />pair A, B, C, D;<br /><br />A = (0,0);<br />B = (5,0);<br />C = (5,2);<br />D = (0,2);<br /><br />draw(A--B--C--D--cycle);<br /><br />draw(Circle(A,0.2));<br />filldraw(Circle(A,0.2),gray(0.7));<br /><br />draw((2.5,1)--(2.5,1.5),Arrow(6));<br />label("$F_{applied}$", (2.5,1.75),N);<br /><br />draw((2.5,1)--(2.5,0.5),Arrow(6));<br />label("$F_g$", (2.5,0.25),S);<br /><br />draw((2.5,1)--(3,1),Arrow(6));<br />label("$F_f$", (3.1,1),E);<br /><br />draw((2.5,1)--(2,1),Arrow(6));<br />// No force opposing applied force in the given scenario as the car is accelerating.<br />//label("$F_{opposing}$", (1.9,1),W); <- Removed this label as it's not applicable.<br /><br />[/asy]<br /><br />* **F_g:** Force of gravity (weight) acting downwards.<br />* **F_applied:** Combined pushing force applied by Om, Rovic, and Dilshan.<br />* **F_f:** Force of friction opposing the motion.<br /><br /><br />**3b. Acceleration:**<br /><br />* **Given:** Net force (F_net) = 25 N, Mass (m) = 1555 kg<br />* **Formula:** F_net = m * a (Newton's Second Law)<br />* **Calculation:** a = F_net / m = 25 N / 1555 kg ≈ 0.016 m/s²<br /><br />**3c. Applied Force:**<br /><br />* **Given:** Coefficient of friction (μ) = 0.320, F_net = 25 N, m = 1555 kg, g = 9.8 m/s² (acceleration due to gravity)<br />* **Formula 1:** F_f = μ * F_N (Force of friction)<br />* **Formula 2:** F_N = m * g (Normal force; equal to weight in this scenario)<br />* **Formula 3:** F_net = F_applied - F_f<br />* **Calculation:**<br /> * F_N = 1555 kg * 9.8 m/s² ≈ 15239 N<br /> * F_f = 0.320 * 15239 N ≈ 4876.5 N<br /> * F_applied = F_net + F_f = 25 N + 4876.5 N ≈ 4901.5 N<br /><br />**3d. Distance:**<br /><br />* **Given:** Initial velocity (v_i) = 0 m/s, time (t) = 3.00 s, acceleration (a) = 0.016 m/s²<br />* **Formula:** d = v_it + (1/2)at² (Equation of motion)<br />* **Calculation:** d = (0 m/s * 3.00 s) + (1/2) * 0.016 m/s² * (3.00 s)² ≈ 0.072 m<br /><br /><br />Therefore, the car could move approximately 0.072 meters or 7.2 centimeters during the 3-second push.<br />