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 =10 N/m. The static friction coefficient between the block and the surface is mμs =0.7 , and the kinetic friction coefficient is mμk =0.4 . Use g = 10 m/s^2 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=
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To solve this problem, we can use the concepts of forces, Newton's laws of motion, and the equations of motion.
(a) To determine the minimum distance x1 required to compress the spring for the block to start moving when released, we need to find the point at which the static friction force is equal to the maximum static friction force. When the block starts moving, the static friction changes to kinetic friction.
The maximum static friction force can be calculated using the formula:
F_max(static friction) = μ_s * Normal force
where μ_s is the static friction coefficient and the Normal force is the force exerted by the surface on the block and is equal to the weight of the block in this case (Normal force = m * g).
So, F_max(static friction) = μ_s * m * g
Next, we need to determine the force exerted by the spring. According to Hooke's Law, the force exerted by a spring is given by:
F_spring = -k * x
where k is the spring constant, and x is the compression or extension of the spring.
Since the static friction force and the force exerted by the spring are in opposite directions, we can equate them:
F_max(static friction) = F_spring
μ_s * m * g = -k * x
Solving for x, we get:
x = (-μ_s * m * g) / k
Plugging in the given values, we can calculate x:
x = (-0.7 * 2 * 10) / 10
x = -1.4 meters
However, x cannot be negative in this context since it represents the compression of the spring. Therefore, we take the absolute value of x:
x1 = |x| = 1.4 meters
So, the minimum distance x1 required to compress the spring for the block to start moving when released is 1.4 meters.
(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 after it starts moving.
Once the block starts moving, the static friction changes to kinetic friction. The force of kinetic friction is given by:
F_kinetic friction = μ_k * Normal force
where μ_k is the kinetic friction coefficient.
The force exerted by the spring is still given by:
F_spring = -k * x
To find the point where the block stops, we need to find the point where the force exerted by the spring is equal to the force of kinetic friction:
F_spring = F_kinetic friction
-k * x = μ_k * Normal force
Substituting the expressions for Normal force and using the fact that the weight of the block is balanced by the normal force:
-k * x = μ_k * m * g
Solving for x, we get:
x = (-μ_k * m * g) / k
Plugging in the given values, we can calculate x:
x = (-0.4 * 2 * 10) / 10
x = -0.8 meters
Again, x cannot be negative in this context, so we take the absolute value:
|x2-x1| = |x2 - (-x1)| = |x2 + x1| = |(0.8 + 1.4)| = 2.2 meters
Therefore, the distance |x2-x1| between the point of release x1 and the point x2 where the block comes to a stop again is 2.2 meters.
(c) To find the time t12 it takes for the block to come to a rest after the release, we need to find the acceleration of the block during this motion.
The net force acting on the block after release is given by:
Net force = Force of kinetic friction + Force exerted by the spring
Net force = -μ_k * m * g - k * x
Acceleration is given by Newton's second law:
Net force = m * acceleration
-m * g * μ_k - k * x = m * acceleration
Substituting the given values, we have:
-2 * 10 * 0.4 - 10 * (-1.4) = 2 * acceleration
Simplifying the equation, we get:
-8 - (-14) = 2 * acceleration
6 = 2 * acceleration
acceleration = 6 / 2 = 3 m/s^2
To find the time t12, we can use the equation of motion:
x = x0 + v0 * t + 0.5 * a * t^2
Since the block starts from rest at x1 with an initial velocity of 0:
x2 = x1 + 0 + 0.5 * a * t^2
Substituting the values:
2.2 = 1.4 + 0 + 0.5 * 3 * t^2
0.8 = 1.5 * t^2
From this equation, we can solve for t:
t^2 = 0.8 / 1.5
t = √(0.8 / 1.5) ≈ 0.565 seconds
Therefore, the time t12 it takes for the block to come to rest after the release is approximately 0.565 seconds.