Question

question 12 (1 point)
an object with a mass m slides down a rough 37° inclined plane where the coefficient of kinetic friction is 0.20. if the plane is 10 m long and the mass starts from rest, what will be its speed at the bottom of the plane?
12 m/s
11 m/s
9.7 m/s
9.3 m/s

Explanation:

Step1: Find the acceleration of the object

First, analyze the forces acting on the object along the inclined plane. The gravitational force component along the plane is \( F_{g\parallel}=mg\sin\theta \), and the frictional force is \( f = \mu_k N \), where the normal force \( N = mg\cos\theta \) (perpendicular to the plane). Using Newton's second law \( F_{net}=ma \), the net force along the plane is \( F_{net}=mg\sin\theta - \mu_k mg\cos\theta \). So the acceleration \( a = g(\sin\theta - \mu_k\cos\theta) \). Given \( \theta = 37^\circ \), \( \mu_k = 0.20 \), and \( g = 9.8\ m/s^2 \). \( \sin37^\circ\approx0.6 \), \( \cos37^\circ\approx0.8 \). Then \( a = 9.8(0.6 - 0.2\times0.8)=9.8(0.6 - 0.16)=9.8\times0.44 = 4.312\ m/s^2 \).

Step2: Use kinematic equation to find final velocity

The object starts from rest (\( u = 0 \)), displacement \( s = 10\ m \), acceleration \( a = 4.312\ m/s^2 \). Use the kinematic equation \( v^2 = u^2 + 2as \). Substitute the values: \( v^2 = 0 + 2\times4.312\times10 = 86.24 \). Then \( v=\sqrt{86.24}\approx9.3\ m/s \).

Answer:

9.3 m/s (the option: 9.3 m/s)