A block with mass 10 kg rests on the horizontal floor that has kinetic friction coefficient of 0.1. A single force that has magnitude of 50 N and an angle of +30degrees with respect to horizontal surface acts on the block so that block slides across the floor with constant acceleration. Determine the speed of the block at the end of 10 m displacement. Select one: a. 7.9m/s b. 7.3m/s c. 8.5m/s d. 7.6m/s 8.2m/s

To solve this problem, we need to use the principles of Newton's second law of motion and the concept of kinetic friction.<br /><br />Given:<br />- Mass of the block, m = 10 kg<br />- Coefficient of kinetic friction, μ = 0.1<br />- Magnitude of the applied force, F = 50 N<br />- Angle of the applied force with respect to the horizontal surface, θ = +30 degrees<br />- Displacement of the block, s = 10 m<br /><br />First, let's calculate the force of kinetic friction acting on the block. The force of kinetic friction is given by the equation:<br /><br />F_kinetic = μ * N<br /><br />where N is the normal force acting on the block. The normal force can be calculated as:<br /><br />N = m * g * cos(θ)<br /><br />where g is the acceleration due to gravity (approximately 9.8 m/s^2).<br /><br />Substituting the given values, we get:<br /><br />N = 10 kg * 9.8 m/s^2 * cos(30 degrees)<br />N ≈ 10 * 9.8 * 0.866<br />N ≈ 84.6 N<br /><br />Now, we can calculate the force of kinetic friction:<br /><br />F_kinetic = 0.1 * 84.6 N<br />F_kinetic ≈ 8.46 N<br /><br />Next, let's calculate the net force acting on the block. The net force is the difference between the applied force and the force of kinetic friction:<br /><br />Net force = Applied force - Force of kinetic friction<br />Net force = 50 N - 8.46 N<br />Net force ≈ 41.54 N<br /><br />According to Newton's second law of motion, the acceleration of the block is given by:<br /><br />a = Net force / m<br />a = 41.54 N / 10 kg<br />a ≈ 4.15 m/s^2<br /><br />Finally, we can calculate the speed of the block at the end of the 10 m displacement using the equation of motion:<br /><br />v^2 = u^2 + 2 * a * s<br /><br />where v is the final speed, u is the initial speed (which is 0 m/s since the block starts from rest), and s is the displacement.<br /><br />Substituting the given values, we get:<br /><br />v^2 = 0 + 2 * 4.15 m/s^2 * 10 m<br />v^2 = 82.6 m^2/s^2<br />v ≈ √82.6 m^2/s^2<br />v ≈ 9.1 m/s<br /><br />Therefore, the speed of the block at the end of the 10 m displacement is approximately 9.1 m/s. None of the given options match this result, so there may be an error in the problem statement or the provided options.