Question

A wooden block of mass m = 9 kg starts from rest on an inclined plane sloped at an angle ? from the horizontal. The block is originally located 5m from the bottom of the plane. If the block, undergoing constant acceleration down the ramp, slides to the bottom in t = 2 s, and ? = 30

Answer 1

The acceleration of the 9 kg wooden block sliding down a 30° inclined plane is calculated using the kinematic equation for displacement, resulting in an acceleration of 2.5 m/s² down the ramp.

Step-by-step explanation:

The question asks to calculate the acceleration of a 9 kg wooden block that starts from rest and slides down an inclined plane with a slope of 30° to the horizontal. The block covers a distance of 5 meters in a time span of 2 seconds. Given these parameters, we can use the kinematic equations to find the block's acceleration.

First, we determine the block's final velocity using the formula v = u + at, where v is the final velocity, u is the initial velocity (0 m/s, in this case), a is the acceleration, and t is the time. However, we do not have the final velocity, so we use another kinematic equation: s = ut + (1/2)at², where s is the displacement (5 m).

Finally, substituting the given values, we have 5 = 0 + (1/2)a(2²). Solving for a, we find that the acceleration of the block is 2.5 m/s² down the ramp.