A 4.10 kg object on a horizontal frictionless surface is attached to a spring with spring constant 840 N/m. The object is displaced from equilibrium 50.0 cm horizontally (in the + direction) and given an initial velocity of 9.9 m/s back toward the equilibrium point at t=0 s
To find the equation of motion and the velocity of the object at any given time, we can use the principles of simple harmonic motion.
First, let's determine the force acting on the object when it is displaced from equilibrium. According to Hooke's Law, the force exerted by the spring is proportional to the displacement from equilibrium:
F = -k * x
Where:
- F is the force exerted by the spring (which is also the net force acting on the object).
- k is the spring constant (840 N/m).
- x is the displacement from equilibrium (50.0 cm or 0.50 m in this case).
Plugging in these values, we get:
F = -(840 N/m) * (0.50 m)
F = -420 N
Since this is the net force acting on the object, we can use Newton's second law to find the acceleration. The formula is:
F = m * a
Where:
- m is the mass of the object (4.10 kg).
- a is the acceleration of the object.
Plugging in the values:
-420 N = (4.10 kg) * a
Solving for a:
a = -420 N / (4.10 kg)
a ≈ -102.44 m/s²
The negative sign indicates that the acceleration is in the opposite direction of the displacement. So, the object will experience an acceleration of approximately -102.44 m/s².
To find the velocity at any given time, we can use the kinematic equation:
v = v₀ + a * t
Where:
- v is the velocity at any given time.
- v₀ is the initial velocity (9.9 m/s in this case).
- a is the acceleration (-102.44 m/s², as calculated above).
- t is the time (in seconds).
Plugging in the values:
v = (9.9 m/s) + (-102.44 m/s²) * t
Therefore, the equation of motion for the object is:
v = 9.9 - 102.44t
This equation allows you to calculate the velocity of the object at any given time after t=0 seconds.
Note: This analysis assumes no external forces (such as friction) are acting on the object.