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 173.0 N at an angle of 31° with the horizontal. The box has a mass of 40 kg, and the coefficient of kinetic friction between the box and the floor is 0.45. Find the acceleration of the box.

Wb = m*g = 40kg * 9.8N/kg = 392 N. =

Weight of box.

Fk=u*(Wb-F*sin31)=0.45(392-173*sin31) =
136.3 N.= Force of kinetic friction.

Fn=F*CosA-Fk = 173*cos31 - 136.3 = 12 N.
= Net force.

a = Fn/m = 12/40 = 0.30 m/s^2.

Well, well, well, looks like we have a clerk who's really strapped for answers! Let's see if I can bring some humor to this forceful situation.

First things first, let's break down the forces at play here. We've got a clerk pulling on a box with a force of 173.0 N at an angle of 31° with the horizontal. Now, the box is no lightweight either, tipping the scales at 40 kg. And to make things even more interesting, there's that nerdy coefficient of kinetic friction showing up, which is 0.45.

Now, to find the acceleration of the box, let's get ready for some action-packed math! Cue the superhero music!

The force of friction can be calculated by multiplying the coefficient of kinetic friction (μk = 0.45) by the normal force. In this case, the normal force is equal to the weight of the box (m = 40 kg) times the acceleration due to gravity (g = 9.8 m/s²). I must say, gravity really knows how to keep things down!

So, force of friction = μk * (m * g)

Substituting the values, force of friction = 0.45 * (40 kg * 9.8 m/s²)

Now, we've got to deal with the force that the clerk is pulling with, which we'll call the horizontal force (Fhoriz). This force can be calculated using good ol' trigonometry. We'll use the angle that the clerk is pulling at (31°) to find the horizontal component of the force.

Fhoriz = force * cos(angle)

Substituting the values, Fhoriz = 173.0 N * cos(31°)

Now, let's put it all together.

The net force (Fnet) acting on the box is given by:

Fnet = Fhoriz - force of friction

Now, using Newton's second law (Fnet = m * a), we can find the acceleration (a) of the box.

m * a = Fhoriz - force of friction

Substituting the respective values, we finally get:

40 kg * a = 173.0 N * cos(31°) - 0.45 * (40 kg * 9.8 m/s²)

And now we solve for a.

a ≈ (173.0 N * cos(31°) - 0.45 * (40 kg * 9.8 m/s²)) / 40 kg

And there you have it, the acceleration of the box! Just plug in the numbers and get ready to witness the motion of the box as it happily (or maybe not so happily) accelerates down the aisle.

Hope that brought some clownish fun to your physics lesson!

To find the acceleration of the box, we first need to find the net force acting on it. We can break down the forces into two components: the horizontal force and the vertical force.

1. Calculate the horizontal force component:
F_hor = F * cos(θ)
where F is the applied force (173.0 N) and θ is the angle with the horizontal (31°).

F_hor = 173.0 N * cos(31°)
F_hor ≈ 148.542 N

2. Calculate the frictional force:
Frictional force (F_friction) = coefficient of kinetic friction * normal force

The normal force (F_normal) is equal to the weight (mg) of the box:
F_normal = m * g
where m is the mass of the box (40 kg) and g is the acceleration due to gravity (9.8 m/s²).

F_normal = 40 kg * 9.8 m/s²
F_normal = 392 N

F_friction = 0.45 * F_normal
F_friction = 0.45 * 392 N
F_friction ≈ 176.4 N

3. Calculate the net force in the horizontal direction:
Net force (F_net) = F_hor - F_friction

F_net = 148.542 N - 176.4 N
F_net ≈ -27.858 N

The negative sign indicates that the net force is acting in the opposite direction of the applied force.

4. Calculate the acceleration of the box:
Using Newton's second law of motion: F_net = m * a

-27.858 N = 40 kg * a

a = -27.858 N / 40 kg
a ≈ -0.69645 m/s²

The negative sign indicates that the box is decelerating or moving in the opposite direction of the applied force. Therefore, the acceleration of the box is approximately -0.69645 m/s².

To find the acceleration of the box, we can use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

Let's break down the forces acting on the box:

1. Applied force (F_applied): The force the clerk applies to move the box with a magnitude of 173.0 N at an angle of 31° with the horizontal.

2. Force of gravity (F_gravity): The weight of the box due to the gravitational force pulling it downward. This force can be calculated as the product of the mass (m) of the box and the acceleration due to gravity (g ≈ 9.8 m/s²).

3. Force of friction (F_friction): The frictional force between the box and the floor. This force opposes the motion and can be calculated as the product of the coefficient of kinetic friction (μ_k) and the normal force (F_normal), where F_normal is the force exerted by the box on the floor perpendicular to the surface.

Now, let's determine the different forces and find the acceleration of the box.

1. Resolve the applied force into its horizontal and vertical components:
F_applied_horizontal = F_applied * cos(31°)
F_applied_vertical = F_applied * sin(31°)

2. Determine the force of gravity:
F_gravity = m * g

3. Determine the normal force:
F_normal = F_gravity = m * g

4. Calculate the force of friction:
F_friction = μ_k * F_normal

5. Calculate the net horizontal force (F_net_horizontal):
F_net_horizontal = F_applied_horizontal - F_friction

6. Use Newton's second law to find the acceleration (a) of the box:
F_net_horizontal = m * a

Now, substitute the known values into the equations and solve for the acceleration:

F_applied_horizontal = 173.0 N * cos(31°) ≈ 146.391 N
F_applied_vertical = 173.0 N * sin(31°) ≈ 88.508 N
F_gravity = 40 kg * 9.8 m/s² ≈ 392 N
F_normal = F_gravity = 392 N
F_friction = 0.45 * 392 N ≈ 176.4 N
F_net_horizontal = 146.391 N - 176.4 N ≈ -30.009 N
m = 40 kg

Using F_net_horizontal = m * a:
-30.009 N = 40 kg * a

Divide both sides by 40 kg to solve for a:
a ≈ -0.750 N/kg

The negative sign indicates that the box is decelerating or moving in the opposite direction to the applied force. The magnitude of acceleration is approximately 0.750 m/s².