Consider an ideal spring that has an unstretched length l_0 = 3.5 m. Assume the spring has a constant k = 36 N/m. Suppose the spring is attached to a mass m = 8 kg that lies on a horizontal frictionless surface. The spring-mass system is compressed a distance of x_0 = 1.9 m from equilibrium and then released with an initial speed v_0 = 5 m/s toward the equilibrium position. How long will it take for the spring to first become completely extended?
http://ocw.mit.edu/courses/physics/8-01t-physics-i-fall-2004/assignments/ps07sol.pdf
A block of mass m = 4 kg slides along a horizontal table when it encounters the free end of a horizontal spring of spring constant k = 14 N/m. The spring is initially on its equilibrium state, defined when its free end is at x=0 . Right before the collision, the block is moving with a speed vi = 5 m/s . There is friction between the block and the surface. The coefficient of friction is given by μ = 0.86 . How far did the spring compress when the block first momentarily comes to rest? take g= 10 m/s2
A block of mass m = 4 kg slides along a horizontal table when it encounters the free end of a horizontal spring of spring constant k = 16 N/m. The spring is initially on its equilibrium state, defined when its free end is at x=0 in the figure. Right before the collision, the block is moving with a speed vi = 9 m/s . There is friction between the block and the surface. The coefficient of friction is given by μ = 0.77 . How far did the spring compress when the block first momentarily comes to rest? take g= 10 m/s2
To find the time it takes for the spring to first become completely extended, we need to analyze the motion of the mass-spring system.
First, we can find the equilibrium position by considering the forces acting on the mass when it is at rest. At equilibrium, the force exerted by the spring is equal to the force of gravity acting on the mass.
The force exerted by the spring can be calculated using Hooke's Law:
F_spring = -k * x
where F_spring is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
The force of gravity acting on the mass is given by:
F_gravity = m * g
where F_gravity is the force of gravity, m is the mass, and g is the acceleration due to gravity.
At equilibrium, these forces are equal, so we have:
-k * x_eq = m * g
Solving for x_eq, we get:
x_eq = -(m * g) / k
Plugging in the values, we have:
x_eq = -(8 kg * 9.8 m/s^2) / 36 N/m ≈ -2.1556 m
Now, we can analyze the motion of the mass-spring system when it is released from the initial position with an initial speed.
The equation of motion for the mass-spring system is:
m * a = -k * x
where m is the mass, a is the acceleration, k is the spring constant, and x is the displacement from the equilibrium position.
We can solve this second-order differential equation to find the position, x(t), as a function of time.
The solution to the equation of motion is:
x(t) = A * cos(ωt - φ)
where A is the amplitude, ω is the angular frequency, and φ is the phase constant.
To determine the amplitude and phase constant, we need to use the initial conditions. Given that the mass is initially compressed a distance of x_0 = 1.9 m from equilibrium and released with an initial speed of v_0 = 5 m/s, we can write:
x(0) = x_0
v(0) = v_0
Differentiating the position equation with respect to time, we get:
v(t) = -A * ω * sin(ωt - φ)
Plugging in the initial conditions, we have:
x(0) = A * cos(-φ) = x_0
v(0) = -A * ω * sin(-φ) = v_0
From the first equation, we can solve for A:
A = x_0 / cos(-φ)
Substituting this into the second equation, we get:
v_0 = -ω * x_0 * tan(-φ)
From these equations, we can solve for the phase constant φ. Once we have φ, we can find ω using the relation between the angular frequency and the spring constant:
ω = sqrt(k / m)
Finally, to find the time it takes for the spring to first become completely extended, we need to find the value of t when x(t) = x_eq.
Setting x(t) = x_eq, we have:
A * cos(ωt - φ) = x_eq
Solving for t, we can find the time it takes for the spring to first become completely extended.