A block of mass m=14 kg slides down a ramp that makes an angle of Theta =15,9^circ degree with the horizontal, accelerating at a=1,84m/s^2 Find the coefficient of kinetic friction between the block and the plane .Take g=9.8m/s2 and round off your result to two decimal places. Yanit: square

To find the coefficient of kinetic friction, we can use the equation:<br /><br />$F_{friction} = \mu_k \cdot F_{normal}$<br /><br />where $F_{friction}$ is the force of friction, $\mu_k$ is the coefficient of kinetic friction, and $F_{normal}$ is the normal force.<br /><br />First, let's calculate the normal force. The normal force is equal to the component of the gravitational force perpendicular to the ramp. The gravitational force acting on the block is given by:<br /><br />$F_{gravity} = m \cdot g$<br /><br />where $m$ is the mass of the block and $g$ is the acceleration due to gravity.<br /><br />The normal force can be calculated as:<br /><br />$F_{normal} = F_{gravity} \cdot \cos(\Theta)$<br /><br />Substituting the given values, we have:<br /><br />$F_{normal} = (14 \, \text{kg}) \cdot (9.8 \, \text{m/s}^2) \cdot \cos(15.9^\circ)$<br /><br />Next, let's calculate the force of friction. The force of friction can be calculated using the equation:<br /><br />$F_{friction} = \mu_k \cdot F_{normal}$<br /><br />We can rearrange this equation to solve for the coefficient of kinetic friction:<br /><br />$\mu_k = \frac{F_{friction}}{F_{normal}}$<br /><br />To find the force of friction, we need to consider the net force acting on the block along the ramp. The net force is the difference between the component of the gravitational force parallel to the ramp and the force of friction. The net force is also equal to the mass of the block multiplied by the acceleration:<br /><br />$F_{net} = m \cdot a$<br /><br />The component of the gravitational force parallel to the ramp is:<br /><br />$F_{parallel} = F_{gravity} \cdot \sin(\Theta)$<br /><br />So, the force of friction is:<br /><br />$F_{friction} = F_{parallel} - F_{net}$<br /><br />Substituting the given values, we have:<br /><br />$F_{net} = (14 \, \text{kg}) \cdot (1.84 \, \text{m/s}^2)$<br /><br />$F_{parallel} = (14 \, \text{kg}) \cdot (9.8 \, \text{m/s}^2) \cdot \sin(15.9^\circ)$<br /><br />Finally, we can calculate the coefficient of kinetic friction:<br /><br />$\mu_k = \frac{F_{parallel} - F_{net}}{F_{normal}}$<br /><br />Substituting the calculated values, we get:<br /><br />$\mu_k = \frac{(14 \, \text{kg}) \cdot (9.8 \, \text{m/s}^2) \cdot \sin(15.9^\circ) - (14 \, \text{kg}) \cdot (1.84 \, \text{m/s}^2)}{(14 \, \text{kg}) \cdot (9.8 \, \text{m/s}^2) \cdot \cos(15.9^\circ)}$<br /><br />Simplifying the expression, we find:<br /><br />$\mu_k \approx 0.11$<br /><br />Therefore, the coefficient of kinetic friction between the block and the plane is approximately 0.11.