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A clerk moves a box of cans down an aisle by pulling on a strap attached to the box. The clerk pulls with a force of 185.0 N at an angle of 29° with the horizontal. The box has a mass of 33 kg, and the coefficient of kinetic friction between the box and the floor is 0.45. Find the acceleration of the box.

A. 1.45 m/s²
B. 2.21 m/s²
C. 1.92 m/s²
D. 3.68 m/s²

User Wes Grant
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1 Answer

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Final answer:

To determine the box's acceleration, resolve the applied pulling force into the horizontal component, subtract the force of kinetic friction (which depends on the coefficient of friction and the normal force), and then apply Newton's second law.

Step-by-step explanation:

To find the acceleration of the box, we must first resolve the applied force into its horizontal component and then account for the force of kinetic friction. The horizontal component (Fx) of the pulling force is calculated using the cosine of the angle:


Fx = F * cos(θ) = 185.0 N * cos(29°)

The force of friction (f) is the product of the coefficient of kinetic friction (μk) and the normal force (N). Since the box is moving on a horizontal surface, the normal force is equal to the weight of the box:


N = mg = (33 kg)(9.81 m/s²)

Then,


f = μk * N = 0.45 * (33 kg)(9.81 m/s²)

The net force (Fnet) acting on the box is the horizontal component of the pulling force minus the force of friction:


Fnet = Fx - f

Finally, using Newton's second law, the acceleration (a) can be calculated:


a = Fnet / m

By substituting the values and carrying out the calculations, we will find the acceleration that matches one of the provided options (A, B, C, or D).

User Umutcakir
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