Final answer:
The tension in the cord supporting the 5-kg mass on the frictionless incline is 49.0 N.
Step-by-step explanation:
The tension in the cord supporting the 5-kg mass on the frictionless incline can be determined by considering the vertical component of the weight of the mass. The weight of the mass is given by the equation W = mg, where m is the mass and g is the gravitational acceleration. In this case, the weight is 5 kg * 9.8 m/s² = 49.0 N.
Since the incline is frictionless, the tension in the cord is equal to the weight of the mass. So the reading on the spring scale would be 49.0 N.