A 120kg crate is pulled across the floor with a constant forceof 500N that is ap

Question

A 120kg crate is pulled across the floor with a constant forceof 500N that is applied at an angle of 25degrees above thehorizontal. For the first 20m of travel the floor is frictionlessand for the next 20m of travel the coefficient of kinetic frictionis 0.25. What is the speed of the crate after 20m of travel? What is the speed of the crate after 40m of travel? A 120kg crate is pulled across the floor with a constant forceof 500N that is applied at an angle of 25degrees above thehorizontal. For the first 20m of travel the floor is frictionlessand for the next 20m of travel the coefficient of kinetic frictionis 0.25. What is the speed of the crate after 20m of travel? What is the speed of the crate after 40m of travel?

Explanation / Answer

mass m = 120 kg force F = 500 N angle = 25 degrees coefficent of friction = 0.25 (a). Force in horizontal direction F ' = F cos = 453.15N Accleration a = F ' / m = 3.776 m / s ^ 2 Speed after 1 st 20 m is v = [ 2aS ]                                        = [ 2* 3.776 * 20 ]                                        = 12.29 m / s (b). next 20 m : --------------- distance S ' = 20 m net force F " = F cos - mg                   = 453.15 - 294 = 159.15 N accleration a ' = F " / m = 1.32625 m / s ^ 2 Speed after 40 m   is V = [ v ^ 2 + 2a'S ' ]                                     = 14.286 m / s

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