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 so 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: 7.9m/s b. 7.3m/s 8.5m/s d 7.6m/s e. 8.2m/s

To determine the speed of the block at the end of a 10 m displacement, we need to consider the forces acting on the block and apply Newton's second law of motion.<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 = +30 degrees<br /><br />First, let's calculate the force of kinetic friction (f_k) acting on the block. The force of kinetic friction is given by the equation:<br /><br />f_k = μ * N<br /><br />where N is the normal force acting on the block. Since the block is on a horizontal surface, the normal force is equal to the weight of the block (mg). Therefore,<br /><br />N = mg<br /><br />Substituting the given values, we have:<br /><br />N = 10 kg * 9.8 m/s^2 = 98 N<br /><br />Now, we can calculate the force of kinetic friction:<br /><br />f_k = 0.1 * 98 N = 9.8 N<br /><br />Next, let's calculate the horizontal component of the applied force (F_x) using trigonometry. The horizontal component of the force is given by:<br /><br />F_x = F * cos(θ)<br /><br />where θ is the angle of the applied force with respect to the horizontal surface. Substituting the given values, we have:<br /><br />F_x = 50 N * cos(30 degrees) = 43.3 N<br /><br />Now, we can apply Newton's second law of motion to find the acceleration (a) of the block. According to Newton's second law, the net force acting on the block is equal to the mass of the block multiplied by the acceleration:<br /><br />F_net = m * a<br /><br />The net force acting on the block is the difference between the horizontal component of the applied force and the force of kinetic friction:<br /><br />F_net = F_x - f_k<br /><br />Substituting the values we calculated earlier, we have:<br /><br />F_net = 43.3 N - 9.8 N = 33.5 N<br /><br />Now, we can solve for the acceleration:<br /><br />33.5 N = 10 kg * a<br /><br />a = 3.35 m/s^2<br /><br />Finally, we can use the equation of motion to find the speed of the block at the end of a 10 m displacement. The equation of motion is:<br /><br />v^2 = u^2 + 2 * a * s<br /><br />where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement. Since the block starts from rest, the initial velocity is zero. Substituting the given values, we have:<br /><br />v^2 = 0 + 2 * 3.35 m/s^2 * 10 m<br /><br />v^2 = 67 m^2/s^2<br /><br />v = √(67 m^2/s^2)<br /><br />v = 8.19 m/s<br /><br />Therefore, the speed of the block at the end of a 10 m displacement is approximately 8.19 m/s. The correct answer is not listed among the options provided.