F = -kx
Force, acceleration and displacement of a pendulum in SHM
A 200g toy robot is attached to a pole by a spring, with a spring constant of 90 N m-1, and made to oscillate horizontally.
(a) What force will act on the robot when it is at its amplitude position of 5 cm from equilibrium?
(b) How fast will the robot accelerate whilst at this amplitude position?
Part (a)
Step 1: Convert amplitude into m
5 cm = 0.05 m
Step 2: Substitute values into the restoring force equation
F = -kx = -(90) x (0.05) = - 4.5 N
Step 3: Explain the answer
A force of 4.5 newtons will act on the robot, trying to pull it back towards the equilibrium position.
Part (b)
Step 1: Convert mass of robot into kg
200 g = 0.2 kg
Step 2: Substitute values into Newton's second law equation:
F = ma
So, = -22.5 m s-2
Step 3: Explain the answer
The train will decelerate at a rate of 22.5 m s-2 when at this amplitude position
Even with this topic you must make sure you convert all quantities into standard SI units
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