An empty crate is given an initial push down a ramp, starting with speed ?v?0, and reaches the bottom with speed ?v? and kinetic energy ?K?. Some books are now placed in the crate, so that the total mass is quadrupled. The coefficient of kinetic fiction is constant and air resistance is negligible. Starting again with ?v?0 at the top of the ramp, what are the speed and kinetic energy at the bottom? Explain the reasoning behind your answers.

Solution 8E Introduction Bot gravitational force and frictional force will act on the crate. We have to calculate the total force acting on the crate and then we can calculate the work done by force. From the given initial velocity we can calculate the initial kinetic energy. Then we can calculate the final kinetic energy. We will repeat the same procedure and then calculate the velocity and kinetic energy for the crate with book. Step 1 The initial speed of the crate is 0 . Suppose the mass of the crate is m. Hence the initial kinetic energy of the crate is K = mv 2. 0 2 0 Now suppose the angle of the ramp with horizontal is . Hence the component of the gravitational force acting in the direction of motion is F = gg sin The normal reaction of the force crate will be N = mg cos So the frictional force is Ff= mg cos Since the crate is coming down, the gravitational force will act towards the direction of motion while the frictional force will act against the direction of motion. Hence the total force acting on the crate is Suppose the length of the ramp is l . Hence the work done by the force on the crate is Hence the final kinetic energy will be Step 2 Suppose the final velocity of the crate will 2e v after the total mass of the crate becomes 4m. So the energy equation will now look like Hence the final velocity of the crate with book will be v .