A student stands on a scale in an elevator that is accelerating at 2.8m/s^2. If the student has a mass of 94 kg, to the nearest newton what is the scale reading?
26.7
To find the scale reading, we first need to understand the forces acting on the student.
When the elevator accelerates, the student experiences two main forces: their weight and the normal force from the scale. The weight of an object is given by the equation:
Weight = mass × acceleration due to gravity
In this case, the weight of the student is:
Weight = 94 kg × 9.8 m/s² (acceleration due to gravity)
Weight = 921.2 N (to the nearest tenth)
Now, let's determine the normal force acting on the student. When the elevator accelerates upwards, the normal force must be greater than the weight to prevent the student from "falling" off the scale.
The equation for the net force acting on a body is:
Net force = mass × acceleration
The net force in this case is the difference between the normal force (upwards) and the weight (downwards), which is:
Net force = normal force - weight
Since the elevator is accelerating upwards, the net force is:
Net force = mass × acceleration of the elevator
Net force = 94 kg × 2.8 m/s²
Net force = 263.2 N (to the nearest tenth)
Since the net force is greater than the weight, we can use the following equation to find the normal force:
Net force = normal force - weight
263.2 N = normal force - 921.2 N
Rearranging the equation to solve for the normal force, we get:
normal force = net force + weight
normal force = 263.2 N + 921.2 N
normal force = 1184.4 N (to the nearest tenth)
Therefore, the scale reading is approximately 1184 N.