You want to move a heavy box with mass 30.0 kg across a carpeted floor. you pull hard on one of the edges of the box at an angle 30∘ above the horizontal with a force of magnitude 240 n, causing the box to move horizontally. the force of friction between the moving box and the floor has magnitude 41.5 n . what is the box's acceleration just after it begins to move?
240 cos 30 - 41.5 = 30a
240*1/2 - 41.5 = 30a
120 - 41.5 = 30a
30a = 78.5
a = 78.5/30 = 2.62 m/s^2
The magnitude of the acceleration of the box is about 5.54 m/s²
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Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.
\(\large {\boxed {F = ma }\)
F = Force ( Newton )
m = Object's Mass ( kg )
a = Acceleration ( m )
Let us now tackle the problem !
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Given:
mass of box = m = 30.0 kg
magnitude of force = F = 240 N
direction of force =
frictional force = f = 41.5 N
Asked:
acceleration of the box = a = ?
Solution:
We will use Newton's Law of Motion to solve this problem as follows:
\(\Sigma F_x = ma\)
\(F_x - f = ma\)
\(F \cos \theta - f = ma\)
\(240 \cos 30^o - 41.5 = 30.0a\)
\(120 \sqrt{3} - 41.5 = 30.0a\)
\(\boxed{a \approx 5.54 \texttt{ m/s}^2}\)
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Grade: High School
Subject: Physics
Chapter: Dynamics