You have an inclined plane of angle 30 degrees. The length of the inclined plane

Question

You have an inclined plane of angle 30 degrees. The length of the inclined plane is 4m. Using only your knowledge of kinetic energy, potential energy, work, and total energy:

A) If uk=0.06, calculate the final velocity of a particle that starts from rest at the top of the plane and ends at the bottom.

B) If the plane is frictionless, calculate the final velocity of a particle that starts from rest at the top of the plane and ends at the bottom.

C) Repeat part A, but you are required to only use the equations of motion rather than energy.

D) Repeat part B, but you are required to only use the equations of motion rather than energy.

Explanation / Answer

The angle of inclination, = 30 deg The length of the inclined plane, l = 4 m a) coefficient of kinetic friction, k = 0.06 initial velocity, u = 0 Since the block moves downwards the acceleration of the block is given by the formula, a = g (sin - kcos)                              = 9.8(sin30 - 0.06cos30)                              = 4.39 m/s^2 We have an equation, v^2 = u^2 + 2as                                             = 0 + 2*4.39*4                                              = 35.12      So the final velocity, v = 5.93 m/s b) Since there is no friction, the formula for the acceleration is given by the formula a = g sin    = 9.8 sin30 = 4.9 m/s^2 Again by using the equation, v^2 = u^2 + 2as                                                       = 0 + 2*4.9*4                                                        = 39.2 So the final velocity, v = 6.26 m/s c) by the formula, a = g (sin - kcos)                              = 9.8(sin30 - 0.06cos30)                              = 4.39 m/s^2 We have an equation, v^2 = u^2 + 2as                                             = 0 + 2*4.39*4                                              = 35.12      So the final velocity, v = 5.93 m/s d) the acceleration is given by a = g sin    = 9.8 sin30 = 4.9 m/s^2 Again by using the equation, v^2 = u^2 + 2as                                                       = 0 + 2*4.9*4                                                        = 39.2 So the final velocity, v = 6.26 m/s                      The angle of inclination, = 30 deg The length of the inclined plane, l = 4 m a) coefficient of kinetic friction, k = 0.06 initial velocity, u = 0 Since the block moves downwards the acceleration of the block is given by the formula, a = g (sin - kcos)                              = 9.8(sin30 - 0.06cos30)                              = 4.39 m/s^2 We have an equation, v^2 = u^2 + 2as                                             = 0 + 2*4.39*4                                              = 35.12      So the final velocity, v = 5.93 m/s b) Since there is no friction, the formula for the acceleration is given by the formula a = g sin    = 9.8 sin30 = 4.9 m/s^2 Again by using the equation, v^2 = u^2 + 2as                                                       = 0 + 2*4.9*4                                                        = 39.2 So the final velocity, v = 6.26 m/s c) by the formula, a = g (sin - kcos)                              = 9.8(sin30 - 0.06cos30)                              = 4.39 m/s^2 We have an equation, v^2 = u^2 + 2as                                             = 0 + 2*4.39*4                                              = 35.12      So the final velocity, v = 5.93 m/s d) by the formula, a = g (sin - kcos)                              = 9.8(sin30 - 0.06cos30)                              = 4.39 m/s^2 We have an equation, v^2 = u^2 + 2as                                             = 0 + 2*4.39*4                                              = 35.12      So the final velocity, v = 5.93 m/s the acceleration is given by a = g sin    = 9.8 sin30 = 4.9 m/s^2 Again by using the equation, v^2 = u^2 + 2as                                                       = 0 + 2*4.9*4                                                        = 39.2 So the final velocity, v = 6.26 m/s                     

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