A 5kg block is released from rest on a smooth plane inclined at an angle of 30o to the horizontal. What is the acceleration down the plane? [g = 10ms-2]

The acceleration down the plane can be found using the formula: a = g sin(theta) where g is the acceleration due to gravity and theta is the angle of inclination of the plane. Given: mass of block, m = 5 kg angle of inclination, theta = 30 degrees acceleration due to gravity, g = 10 m/s^2 We can first find the component of gravity acting down the plane: force due to gravity along the plane = m g sin(theta) where sin(theta) = sin(30 degrees) = 0.5 force due to gravity along the plane = 5 x 10 x 0.5 = 25 N Since there are no other forces acting on the block, the force along the plane is equal to the force causing the acceleration of the block down the plane. Using Newton's second law of motion, we can find the acceleration down the plane: force along the plane = m a where a is the acceleration down the plane. Substituting the values, we get: 25 N = 5 kg x a a = 25 / 5 = 5 m/s^2 Therefore, the acceleration down the plane is 5 m/s^2. Looking at the options given, we can see that the correct answer is the first option, 5.0 ms^-2.