a 24.3 n box feels a net force forward of 11.4 n. if it started from rest, how far does the box move in 1.00 s?
(ignore friction)

8.70 m
6.44 m
2.30 m
3.04 m

Question

a 24.3 n box feels a net force forward of 11.4 n. if it started from rest, how far does the box move in 1.00 s?
(ignore friction)

8.70 m
6.44 m
2.30 m
3.04 m

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Newtons Second Law", "Kinematic Equations" ], "new_concepts": [ "Force and Motion Integration" ], "current_concepts": [ "Newtons Second Law", "Kinematic Equations", "Force and Motion Integration" ] } ## Calculate the mass of the box Using the *Newtons Second Law* knowledge point $$ \begin{aligned} &F_g = 24.3\text{ N},\quad g = 9.80\text{ m/s}^2\ &m = \frac{F_g}{g} = \frac{24.3}{9.80} \approx 2.4796\text{ kg} \end{aligned} $$ ## Calculate the acceleration of the box Using the *Newtons Second Law* knowledge point $$ \begin{aligned} &F_{\text{net}} = 11.4\text{ N}\ &a = \frac{F_{\text{net}}}{m} = \frac{11.4}{2.4796} \approx 4.5975\text{ m/s}^2 \end{aligned} $$ ## Calculate the distance traveled Using the *Kinematic Equations* knowledge point $$ \begin{aligned} &v_0 = 0\text{ m/s},\quad t = 1.00\text{ s}\ &d = v_0 t + \frac{1}{2} a t^2 = 0 + \frac{1}{2} (4.5975)(1.00)^2 \approx 2.30\text{ m} \end{aligned} $$

(A) 8.70 m

(B) 6.44 m

(C) 2.30 m

(D) 3.04 m

{ "subject": "Natural Science", "question_type": "Multiple Choice", "knowledge_point": [ "Natural Science", "Physics", "Force and Motion Integration" ] }

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QUESTION IMAGE

Question

Explanation:

Calculate the mass of the box

Calculate the acceleration of the box

Calculate the distance traveled

Answer: