Question

an ice skater applies a horizontal force to a 20.-kilogram block on frictionless, level ice, causing the block to accelerate uniformly at 1.4 meters per second2 to the right. after the skater stops pushing the block, it slides onto a region of ice that is covered with a thin layer of sand. the coefficient of kinetic friction between the block and the sand-covered ice is 0.28. calculate the magnitude of the force applied to the block by the skater. [show all work, including the equation and substitution with units.]

Answer 1

The only vertical forces are weight and normal force, and they balance since the surface is horizontal. The horizontal forces are the applied force (uppercase F) in the direction the block slides and the frictional force (lowercase f) in the opposite direction.

Apply Newton's 2nd Law in the horizontal direction:
ΣF = ma
F - f = ma
where f = µmg

F - µmg = ma
F = m(a +µg)
F = (20 kg)(1.4 m/s² + 0.28(9.8 m/s²)

F = 83 N