Question

Block 1 (5 kg) is located on the surface of a table. A hand pushes horizontally to the right on block 1 with a normal force of 85 N. The coefficient of kinetic friction between the block and the surface equals 0.8. On a sheet of paper, draw the free body diagram for block 1 using the two-subscript notation from class. After completing the free body diagram, enter below each force and its x & y-components. Remember that the x-component is the "i" component and the y-component is the "j" component. FORCES on BLOCK 1 Weight force on block 1 by Earth W1E = i + j N Remember that W1E points DOWN (negative y-value) with a magnitude of m1 * g. Normal force on block 1 by Surface N1S = i + j N Since the block does not move vertically, the upward N1S force must cancel the downward W1E force. Normal force on block 1 by Hand N1H = i + j N The magnitude of N1H is given in the problem and points right (positive x-value). Frictional force on block 1 by Surface f1S = i + j N The kinetic friction on block 1 by the surface (f1S) equals the kinetic coefficient of friction times the normal force between block 1 and the surface (N1S). In this particular case with another external force acting on the block (N1H), the kinetic friction opposes the direction of motion. Incorrect. Tries 1/2 Previous Tries What is the acceleration a of block 1? a = i + j m/s2

Answer 1