You place a 10 kg block on a ramp with an angle of 20 degrees. You push the block up the ramp giving it an initial velocity of 15 m/s. If the coefficient of friction between the block and the ramp is 0.4, find the total distance the block travels before it turns around and slides back down the ramp.

L = 15.97 m

Explanation:

Given:-

- The mass of the block, m = 10 kg

- The inclination of ramp, θ = 20°

- The initial speed, Vi = 15 m/s

- The coefficient of friction u = 0.4

Find:-

find the total distance the block travels before it turns around and slides back down the ramp.

Solution:-

- The total distance travelled by the block up the ramp is defined when all the kinetic energy is converted into potential energy and work is done against the friction. Final velocity V2 = 0.

- Develop a free body diagram of the block. Resolve the weight "W" of the block normal to the surface of ramp. Then apply equilibrium condition for the block in the direction normal to the surface:

                                N - W*cos( θ ) = 0

Where, N : The contact force between block and ramp.

                                N = m*g*cos ( θ )

- The friction force (Ff) is defined as:

                               Ff = u*N

                               Ff = u*m*g*cos ( θ )

- Apply the work-energy principle for the block which travels a distance of "L" up the ramp:

                               K.E i = P.E f + Work done against friction

Where,  K.E i = 0.5*m*Vi^2

             P.E f = m*g*L*sin( θ )

             Work done = Ff*L

- Evaluate "L":

                        0.5*m*Vi^2 = m*g*L*sin( θ ) + u*m*g*cos ( θ )*L

                        0.5*Vi^2 = g*L*sin( θ ) + u*g*cos ( θ )*L

                        0.5*Vi^2 = L [ g*sin( θ ) + u*g*cos ( θ ) ]

                        L = 0.5*Vi^2 / [ g*sin( θ ) + u*g*cos ( θ ) ]

                        L = 0.5*15^2 / [ 9.81*sin( 20 ) + 0.4*9.81*cos ( 20 ) ]

                        L = 15.97 m

does anyone know or what?