A block of mass m=2 kg on a horizontal surface is connected to a spring connected to a wall (see figure).
The spring has a spring constant k= 12 N/m. The static friction coefficient between the block and the surface is μs= 0.6 , and the kinetic friction coefficient is μk= 0.2 . Use g=10 m/s2 for the gravitational acceleration.
(a) The spring is initially uncompressed and the block is at position x=0. What is the minimum distance x1 we have to compress the spring for the block to start moving when released? (in meters)
x1=
(b) Find the distance |x2−x1| between the point of release x1 found in (a), and the point x2 where the block will come to a stop again. (in meters)
|x2−x1|=
(c) What time t12 does it take the block to come to a rest after the release? (i.e., the time of travel between points x1 and x2; in seconds)
t12=
(d) What will happen after the block has come to a rest at point x2?
1.The block will move back towards x1, and it will oscillate with constant frequency and exponentially decreasing amplitude.
2.The block will move back towards x1, and it will oscillate while decreasing both frequency and amplitude.
3.The block will start moving back towards x1, and it will come to a final halt before reaching it.
4.The block will stay at its resting position x2.
5.The answer depends on whether x2>0 or x2<0
plz formulas and answers)))
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To solve this problem, we need to use Newton's laws of motion, the equations governing spring motion, and the equations for friction.
(a) To find the minimum distance x1 to compress the spring for the block to start moving, we need to consider the forces acting on the block. When the block is at rest, the static friction between the block and the surface prevents it from moving. The force exerted by the spring, which is given by Hooke's Law, can balance this static friction force. The equation for Hooke's Law is F = -kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
Since the block is not moving, the maximum static friction force is equal to μs * mg, where μs is the static friction coefficient and mg is the weight of the block. Therefore, to find the minimum distance x1, we need to set the static friction force equal to the force exerted by the spring:
μs * mg = -kx1
Substituting the given values, μs = 0.6, m = 2 kg, g = 10 m/s^2, and k = 12 N/m, we can solve for x1:
0.6 * 2 * 10 = -12 * x1
12 = 12 * x1
x1 = 1 meter
Therefore, the minimum distance x1 required to compress the spring for the block to start moving is 1 meter.
(b) To find the distance |x2−x1| between the point of release x1 and the point x2 where the block will come to a stop again, we need to consider the forces acting on the block. When the block is released, the static friction force is replaced by kinetic friction force. The kinetic friction force is given by Fk = μk * mg, where μk is the kinetic friction coefficient.
The force exerted by the spring will gradually decrease as the block moves. At point x2, the spring force will be zero, and the only force acting on the block will be the kinetic friction force. This friction force will bring the block to a stop.
To find x2, we can use the work-energy principle. The work done by the spring force is equal to the work done by kinetic friction force. The work done by the spring force is given by (1/2)kx1^2, and the work done by kinetic friction force is given by Fk * |x2 - x1|. Equating these two:
(1/2)kx1^2 = Fk * |x2 - x1|
Substituting the given values, μk = 0.2, m = 2 kg, g = 10 m/s^2, and k = 12 N/m, we can solve for |x2 - x1|:
(1/2) * 12 * 1^2 = 0.2 * 2 * 10 * |x2 - 1|
6 = 4 * |x2 - 1|
|x2 - 1| = 1.5 meters
Therefore, the distance |x2−x1| between the point of release x1 and the point x2 where the block will come to a stop again is 1.5 meters.
(c) To find the time t12 it takes the block to come to a rest, we can use the equation of motion for a block with kinetic friction:
x = x1 + v1 * t + (1/2) * (Fk / m) * t^2,
where x is the displacement from x1, v1 is the initial velocity, and Fk / m is the acceleration due to kinetic friction.
The block comes to a rest at point x2, so we can set x = x2 - x1, v1 = 0, and solve for t:
x2 - x1 = (1/2) * (Fk / m) * t^2,
1.5 = (1/2) * (0.2 * 2 * 10 / 2) * t^2,
1.5 = 0.2 * t^2,
t^2 = 7.5,
t = √7.5 ≈ 2.74 seconds.
Therefore, the time t12 it takes the block to come to rest after the release is approximately 2.74 seconds.
(d) After the block has come to a rest at point x2, it will start moving back towards x1. Due to the restoring force of the spring, it will oscillate between x1 and x2. The correct answer is 1. The block will move back towards x1, and it will oscillate with a constant frequency and exponentially decreasing amplitude.
Note: The above solutions assume there is no damping in the system. If there were damping, the block's motion would eventually stop due to the dissipation of mechanical energy.