Q.) 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= 18 N/m. The static friction coefficient between the block and the surface is μs= 0.7 , and the kinetic friction coefficient is μk= 0.3 . 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)
(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)
(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)
To solve this problem, we need to consider the forces acting on the block and the equations governing its motion.
First, let's analyze the forces acting on the block when it's at rest:
1. Gravitational force (Fg): This force acts vertically downwards and is given by Fg = m*g, where m is the mass of the block and g is the acceleration due to gravity. In this case, Fg = 2 kg * 10 m/s^2 = 20 N.
2. Normal force (Fn): This force balances out the gravitational force and acts perpendicular to the surface. Since the block is at rest, Fn = Fg = 20 N.
3. Static friction force (Fs): This force opposes the motion of the block and acts parallel to the surface. The magnitude of static friction is given by Fs = μs * Fn. Here, μs = 0.7. So, Fs = 0.7 * 20 N = 14 N.
Since the block is at rest, the static friction force must balance out the force applied to the block by the compressed spring. In other words, Fs = k * x1, where k is the spring constant and x1 is the compressed distance.
(a) To find the minimum distance x1 we need to compress the spring, we equate the static friction force to the force applied by the spring:
Fs = k * x1
14 N = 18 N/m * x1
x1 = 14/18 m
≈ 0.778 m
So, the minimum distance x1 we have to compress the spring for the block to start moving when released is approximately 0.778 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 when it's in motion:
1. Gravitational force (Fg): Same as before.
2. Kinetic friction force (Fk): Once the block starts moving, the kinetic friction force comes into play. Its magnitude is given by Fk = μk * Fn. Here, μk = 0.3.
3. Force applied by the compressed spring (Fs): This force pushes the block forward and is given by Fs = k * x, where x is the distance traveled by the block.
When the block comes to a stop at point x2, the force applied by the spring (Fs = k * x2) and the kinetic friction force (Fk = μk * Fn) balance out each other. So, we can write:
k * x2 = μk * Fn
18 N/m * x2 = 0.3 * 20 N
x2 = (0.3 * 20 N) / (18 N/m)
≈ 0.333 m
To find |x2 - x1|, we subtract the compressed distance x1 from the distance x2:
|x2 - x1| = |0.333 m - 0.778 m|
≈ 0.445 m
So, the distance |x2 - x1| between the point of release x1 and the point x2 where the block will come to a stop again is approximately 0.445 meters.
(c) To find the time t12 it takes the block to come to rest after the release, we can use the equation of motion for constant acceleration:
x = x1 + v1 * t + (1/2) * a * t^2
At point x2, the block comes to a stop, so its final velocity (v2) is zero. We can write:
x2 = x1 + v1 * t12 + (1/2) * a * t12^2
Since the block starts from rest, its initial velocity (v1) is zero. The acceleration (a) is given by:
a = (Fs - Fk) / m
Substituting the known values, we have:
a = (k * x1 - μk * Fn) / m
= (18 N/m * 0.778 m - 0.3 * 20 N) / 2 kg
Using this value of acceleration and the given values of x1 and x2, we can solve the equation for t12.
I will now calculate the value of t12.
To solve this problem, we will use the concepts of force, friction, and energy.
(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. The force exerted by the spring when it is compressed is given by Hooke's Law:
Fs = -kx
where Fs is the force exerted by the spring, k is the spring constant, and x is the compression distance.
The maximum static friction force that can be exerted between the block and the surface is given by:
Fs_max = μs * N
where μs is the static friction coefficient and N is the normal force. The normal force in this case is equal to the weight of the block, N = mg, where m is the mass of the block and g is the gravitational acceleration.
We know that the block will start moving when the force exerted by the spring is equal to the maximum static friction force:
-kx1 = μs * mg
Plugging in the given values:
-18x1 = 0.7 * 2 * 10
Simplifying the equation:
-18x1 = 14
Dividing both sides by -18:
x1 = -14/18
Taking the absolute value:
x1 = 0.778 m
So, the minimum distance we have to compress the spring for the block to start moving is 0.778 meters.
(b) To find the distance |x2 - x1| between the point of release x1 and the point x2 where the block comes to a stop, we need to consider the energy conservation principle. The total mechanical energy of the system (block + spring) is conserved:
K1 + U1 = K2 + U2
where K1 and K2 are the kinetic energies at points x1 and x2, and U1 and U2 are the potential energies at points x1 and x2.
At point x1, the block is at rest, so its kinetic energy is zero: K1 = 0.
The potential energy at point x1 is given by the elastic potential energy stored in the spring, U1 = 1/2 * k * x1^2.
At point x2, the block comes to a stop, so its kinetic energy is zero: K2 = 0.
The potential energy at point x2 is also given by the elastic potential energy stored in the spring, U2 = 1/2 * k * x2^2.
Since the block comes to a stop at point x2, the total energy is equal to the potential energy at point x2.
0 + 1/2 * k * x1^2 = 1/2 * k * x2^2
Plugging in the given values:
0 + 1/2 * 18 * 0.778^2 = 1/2 * 18 * x2^2
Simplifying the equation:
0.5 * 18 * 0.778^2 = 0.5 * 18 * x2^2
Dividing both sides by 0.5 * 18:
0.778^2 = x2^2
Taking the square root of both sides:
x2 = ±0.778
Since we're interested in the distance, we take the absolute value:
x2 = 0.778 m
The distance |x2 - x1| is equal to the absolute difference between x2 and x1:
|x2 - x1| = |0.778 - 0.778| = 0
So, the distance between the point of release x1 and the point x2 where the block will come to a stop is 0 meters.
(c) To find the time t12 it takes for the block to come to a rest after the release, we need to consider the motion of the block and the equation of motion.
The equation of motion for the block can be written as:
F_net = m * a
where F_net is the net force acting on the block, m is the mass of the block, and a is the acceleration.
The total net force acting on the block is the force exerted by the spring minus the force of kinetic friction:
F_net = Fs - Fk
where Fs is the force exerted by the spring and Fk is the force of kinetic friction.
The force exerted by the spring is given by Hooke's Law:
Fs = -k * x
where k is the spring constant, and x is the displacement of the block from its equilibrium position.
The force of kinetic friction is given by:
Fk = μk * N
where μk is the kinetic friction coefficient and N is the normal force.
At point x2, the block comes to a stop, so the net force is equal to zero:
0 = -k * x2 - μk * N
Since N = mg, we can rewrite the equation as:
0 = -k * x2 - μk * mg
Plugging in the given values:
0 = -18 * 0.778 - 0.3 * 2 * 10
Simplifying the equation:
0 = -14.004 - 6
0 = -20.004
The equation does not have a solution, which means that the block will not come to a rest after the release.