A small block of mass 200 g is situated at the top of a frictionless incline which is 10 m long and 3.2 m high. Determine the speed of the block when it reaches the ground if it falls off the incline and drops vertically to the ground. Take g = 10 m/s ^2 .

Step-by-step solution

By the principle of conservation of mechanical energy, the potential energy of the block at the top is converted entirely into kinetic energy when it reaches the ground. mgh = 1 2 mv^2 v = 2gh Given that the height of the incline is h = 3.2 m and g = 10 m/s ^2 , we substitute these values into the equation: v = 2 10 3.2 v = 64 = 8.0 m/s The mass of the block and the length of the incline are extraneous information.