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 188.0 N at an angle of 26° with the horizontal. The box has a mass of 30 kg, and the coefficient of kinetic friction between the box and the floor is 0.45. Find the acceleration of the box.
in m/s2 down the aisle
remember the horizontal component of his pulling is the pulling force
hte upward component of his pulling just reduces the normal force, thence reduces friction.
net force=ma
horizontal force-mu*normalforce=ma
solve for a.
To find the acceleration of the box, we first need to calculate the net force acting on the box.
First, let's find the horizontal component of the force exerted by the clerk. The horizontal component can be found using the formula:
F_horizontal = F * cos(theta)
where F is the magnitude of the force and theta is the angle with the horizontal.
Substituting the values given in the problem:
F_horizontal = 188.0 N * cos(26°)
F_horizontal ≈ 169.53 N
Next, let's calculate the frictional force acting on the box. The frictional force can be found using the formula:
Frictional force = coefficient of kinetic friction * normal force
The normal force is the force exerted by the floor on the box, which is equal to the weight of the box (mass * gravity).
Normal force = mass * gravity
Normal force = 30 kg * 9.8 m/s^2
Normal force ≈ 294 N
Frictional force = 0.45 * 294 N
Frictional force ≈ 132.3 N
Now, to find the net force acting on the box, we subtract the frictional force from the horizontal force:
Net force = F_horizontal - Frictional force
Net force = 169.53 N - 132.3 N
Net force ≈ 37.23 N
Lastly, we can use Newton's second law of motion to calculate the acceleration of the box:
Net force = mass * acceleration
37.23 N = 30 kg * acceleration
acceleration = 37.23 N / 30 kg
acceleration ≈ 1.24 m/s^2
Therefore, the acceleration of the box down the aisle is approximately 1.24 m/s^2.