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 187.0 N at an angle of 25.0◦ with the horizontal. The box has a mass of 36.0 kg, and the coefficient of kinetic
friction between the box and floor is 0.410. The acceleration due to gravity is 9.81 m/s^2
What is the acceleration of the box?
When i tried to solve for it i did:
Sin(25)*187+(36*9.81) = 432.19
0.410(432.19) = 177.198
187*COS(25)-177.198= -7.71819
A = -7.71819/36 = -0.214394
The answer i submitted was wrong. i wonder if it was because i got a negative number as the Kinetic Force. Please review and help solve this.
Huh?
Sin(25)*187+(36*9.81) = 432.19
the clerk is pulling UP not down
-Sin(25)*187+(36*9.81) =274.1 = normal force down on floor
274.1 * .410 = 112
so
187 cos 25 - 112 = 36 a
To solve for the acceleration of the box, you need to consider the forces acting on it. Let's break it down step-by-step:
Step 1: Resolve the applied force into its components.
The applied force can be divided into two components: one parallel to the floor and one perpendicular to the floor.
The component parallel to the floor can be found using cosine:
F_parallel = F_applied * cos(angle)
F_parallel = 187.0 N * cos(25.0°)
F_parallel = 170.464 N
Step 2: Determine the frictional force.
The frictional force can be found using the coefficient of kinetic friction and the normal force acting on the box.
The normal force (Fn) is equal to the weight of the box (mass * gravity):
Fn = m * g
Fn = 36.0 kg * 9.81 m/s^2
Fn = 352.08 N
The frictional force (F_friction) is equal to the coefficient of kinetic friction multiplied by the normal force:
F_friction = μ * Fn
F_friction = 0.410 * 352.08 N
F_friction = 144.07168 N
Step 3: Calculate the net force.
The net force acting on the box is equal to the applied force minus the frictional force:
F_net = F_parallel - F_friction
F_net = 170.464 N - 144.07168 N
F_net = 26.39232 N
Step 4: Determine the acceleration of the box.
Using Newton's second law, F = m * a, you can find the acceleration:
F_net = m * a
a = F_net / m
a = 26.39232 N / 36.0 kg
a ≈ 0.7334542 m/s^2
Therefore, the acceleration of the box is approximately 0.733 m/s^2.
To calculate the acceleration of the box, we can use Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
To find the net force, we need to consider the forces acting on the box. In this case, the forces include the pulling force by the clerk and the force of friction.
1. The horizontal component of the pulling force can be found using the equation F_horizontal = F * cos(theta), where F is the pulling force (187.0 N) and theta is the angle with the horizontal (25.0°). Plugging in the values, we get F_horizontal = 187.0 N * cos(25.0°) ≈ 168.865 N.
2. The force of friction can be calculated as F_friction = coefficient of friction * normal force. The normal force is equal to the weight of the box, which can be calculated as mass * acceleration due to gravity. Plugging in the values, we get F_friction = 0.410 * (36.0 kg * 9.81 m/s^2) ≈ 143.019 N.
Now, let's calculate the net force:
Net force = F_horizontal - F_friction = 168.865 N - 143.019 N ≈ 25.846 N.
Finally, we can calculate the acceleration of the box using Newton's second law:
Net force = mass * acceleration.
25.846 N = 36.0 kg * acceleration.
Dividing both sides of the equation by 36.0 kg, we find:
acceleration = 25.846 N / 36.0 kg ≈ 0.718 m/s^2.
Therefore, the acceleration of the box is approximately 0.718 m/s^2.