Question

Phenomenon: Cutting the grass You push a 20 kg lawn mower, exerting a 100 N force on it. You push 37° below the horizontal. The effective coefficient of kinetic friction between the grass and mower is uk=0.60. The Big Question: What is the acceleration of the lawn mower? Part a) Draw a picture of the scenario being sure to define a coordinate system as well as draw and label arrows to represent relevant physical quantities. Part b) Draw a force diagram for the lawn mower. How many forces are acting on the lawn mower? How do you know? Is this consistent with the number of solid lines on your force diagram? (Hint: the grass on lawn mower should have a component parallel to the mower's motion and a component perpendicular to the mower's motion drawn in dashed lines. Part c) Write two mathematical representations for the mower, one for the direction parallel (i.e. the x direction) and one for the direction perpendicular (i.e. the y direction) to the mower's motion. What is a ? How do you know? Part d) What is Fyou on Mower ? Fyou on Mower 1 ? Substitute these values into corresponding mathematical representations in part (C). To determine the parallel component of the force of the grass on the mower, use: Fsurface on System | = MkFsurface on System (where the "surface" is the grass and the “system” is the mower). use the mathematical representation for the mower in the perpendicular direction from part (c) to find Forass on Mower1. Use this to determine FGrass on Mower - Part e) Use your answers from parts (a)-(d) to determine a . Be sure to answer in the form of an argument and address any assumptions made.

Answer 1

3 0 37 a FSinceas there torcen the lawn maver are 3 ading Nomel and frrchon yore are cempren the porce appred by the Gress ng 9-axis ma 1 = t-4 FCUO) ma FL-FSIN (37°) Syfen On For ce sucpe trA USing N m - Sin (09) tmg lTsn (3)+mg Fin=Fca(3a) = 9.86N F&-FSin (33) - 153.1N Jyspen =60 (9N F=9.JCN Its no pasible fo move the lawn mower Fsye /57.7N a=O