When the spring is extended by 44.5 cm - 34.0 cm = 10.5 cm = 0.105 m, it exerts a restoring force with magnitude R such that the net force on the mass is
∑ F = R - mg = 0
where mg = weight of the mass = (7.00 kg) g = 68.6 N.
It follows that R = 68.6 N, and by Hooke's law, the spring constant is k such that
k (0.105 m) = 68.6 N ⇒ k = (68.6 N) / (0.105 m) ≈ 653 N/m
Meters Micrometers centimeters millimeters
Magnitude of acceleration
Explanation:
We know that acceleration can increase depending in the force applied on an object, any object with a greater mass will apply a greater force. F = M(a).
Answer:
Toward the centre of the circular path
Explanation:
The can is moved in a circular path: this means that it is moving by circular motion (uniform circular motion if its tangential speed is constant).
In order to keep a circular motion, an object must have a force that pushes it towards the centre of the circular trajectory: this force is called centripetal force, and its magnitude is given by

where m is the mass of the object, v its tangential speed, r the radius of the trajectory. This force always points towards the centre of the circular path.