Question

As shown in the figure below, a box of mass m = 69.0 kg (initially at rest) is pushed a distance d = 73.0 m across a rough warehouse floor by an applied force of FA = 216 N directed at an angle of 30.0° below the horizontal. The coefficient of kinetic friction between the floor and the box is 0.100. Determine the following. (For parts (a) through (d), give your answer to the nearest multiple of 10.)

Answer 1

The physics problem requires calculating the net force exerted on the box, the work done by the applied force over a distance, and the final velocity of the box, taking into account the angle of the force and kinetic friction.

Step-by-step explanation:

Calculating Net Force, Work Done, and Final Velocity

To solve this physics problem involving kinetic friction and an applied force, we must apply concepts of mechanics and work-energy principles. The student asked to determine various results for a 69.0 kg box being pushed across a rough floor.

First, we resolve the applied force into horizontal and vertical components and subtract the force of kinetic friction to find the net force on the box. The force of kinetic friction is given by µkmgcos(θ), where µk is the coefficient of kinetic friction, m is the mass of the box, g is acceleration due to gravity, and θ is the angle of the applied force. To calculate the work done by the force over the distance, we use the formula Work = Force × Distance × cos(θ).

Finally, we use the work-energy theorem to find the final velocity of the box after being pushed the given distance, where the work done by the applied force is equal to the change in kinetic energy of the box.

Answer 2

The problem is analyzing the net force on a box being pushed on a rough surface. The normal force and frictional force are calculated, and then subtracted from the horizontal component of the applied force to get the net force.

Step-by-step explanation:

In this problem, a box with mass m = 69.0 kg is being pushed across a floor with friction by a force FA = 216 N at an angle of 30.0° below the horizontal. The coefficient of kinetic friction between the box and the floor is 0.100.

Firstly, we need to calculate the normal force. Normal force (N) usually equals to the weight of the object, so N = m*g, where g is the gravitational force which is approximately 9.8 m/s². The weight would be W = mg = 69.0 kg * 9.8 m/s² = 676.2 N.

The frictional force is the coefficient of kinetic friction, which is 0.100, times the normal force N. Hence, frictional force = µk * N = 0.100 * 676.2 N = 67.62 N.

Next, we determine the horizontal force, which is the force FA (216 N) times the cosine of 30°, so FA cos(30.0°) = 216 N * cos(30.0°) = 187.1 N. The net force on the box caused by FA and the frictional force is then the horizontal force minus the frictional force, i.e., 187.1 N - 67.62 N = 119.48 N.

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