2. A 20.0-kg block is connected to a 30.0-kg block by a
string that passes over a light frictionless pulley. The 30.0-
kg block is connected to a spring that has negligible mass
and a force constant of 250 N/m, as shown in the Figure.
The spring is upstretched when the system is as shown in
the figure, and the incline is frictionless. The 20.0-kg block
is pulled 20.0 cm down the incline (so that the 30.0-kg block
is 40.0 cm above the floor) and released from rest. Find the
speed of each block when the 30.0-kg block is 20.0 cm
above the floor (that is, when the spring is upstretched)
To find the speed of each block when the 30.0 kg block is 20.0 cm above the floor, we can use the principle of conservation of mechanical energy.
Step 1: Find the potential energy of the 20.0 kg block when it is pulled 20.0 cm down the incline.
Using the formula for gravitational potential energy, we have
Potential energy (20 kg block) = mass * gravity * height
Potential energy (20 kg block) = 20.0 kg * 9.8 m/s^2 * 0.20 m
Step 2: Find the potential energy of the 30.0 kg block when it is 40.0 cm above the floor.
Potential energy (30 kg block) = mass * gravity * height
Potential energy (30 kg block) = 30.0 kg * 9.8 m/s^2 * 0.40 m
Step 3: Find the potential energy stored in the spring when it is upstretched.
Potential energy (spring) = (1/2) * force constant * displacement^2
Potential energy (spring) = (1/2) * 250 N/m * (0.20 m)^2
Step 4: At this point, the system has reached its maximum potential energy, and the energy will convert to kinetic energy when the spring is released. So, the total potential energy at the start is equal to the total kinetic energy at the end.
Potential energy (20 kg block) + Potential energy (30 kg block) + Potential energy (spring) = Kinetic energy (20 kg block) + Kinetic energy (30 kg block)
Step 5: Solve for the speed of each block.
We need to use the equation for kinetic energy, which is
Kinetic energy = (1/2) * mass * velocity^2
Let's assume the velocity of the 20 kg block is v1 and the velocity of the 30 kg block is v2.
Setting up the energy equation, we have
20.0 kg * 9.8 m/s^2 * 0.20 m + 30.0 kg * 9.8 m/s^2 * 0.40 m + (1/2) * 250 N/m * (0.20 m)^2 = (1/2) * 20.0 kg * v1^2 + (1/2) * 30.0 kg * v2^2
Step 6: Rearrange the equation and solve for v1 and v2.
(1/2) * 20.0 kg * v1^2 + (1/2) * 30.0 kg * v2^2 = 20.0 kg * 9.8 m/s^2 * 0.20 m + 30.0 kg * 9.8 m/s^2 * 0.40 m + (1/2) * 250 N/m * (0.20 m)^2
Simplifying the equation, we get
10.0 kg * v1^2 + 15.0 kg * v2^2 = 20.0 kg * 9.8 m/s^2 * 0.20 m + 30.0 kg * 9.8 m/s^2 * 0.40 m + 5 J
Now, we can solve for v1 and v2 by substituting the values calculated in Step 1, Step 2, and Step 3, and then solving the equation.
To solve this problem, we need to consider the forces acting on each block and use Newton's laws of motion.
Let's analyze Block 1 (20.0 kg block) first. It is being pulled vertically downward with a force equal to its weight, mg. The force of gravity acting on Block 1 can be calculated as:
Force1 = mass1 * gravity = 20.0 kg * 9.8 m/s^2 = 196.0 N
Now, looking at Block 2 (30.0 kg block), it is being pulled upward by the tension in the string, T, and it is also subjected to the force from the spring, Fs. The force of gravity acting on Block 2 is given by:
Force2 = mass2 * gravity = 30.0 kg * 9.8 m/s^2 = 294.0 N
Since the pulley is light and frictionless, the tensions in the string on both sides of the pulley will be equal. Therefore, Tension (T) in the string can be considered as the force acting on Block 1.
Now, let's consider the force from the spring (Fs) acting on Block 2. Since the spring is being stretched, the force from the spring is given by Hooke's law:
Fs = -k * x
Where k is the force constant of the spring (250 N/m) and x is the displacement of Block 2 from its equilibrium position.
In this scenario, when Block 1 is pulled 20.0 cm down the incline, Block 2 is 40.0 cm above the floor. So x in this case is (40.0 cm - 20.0 cm) = 20.0 cm = 0.20 m.
Plugging the values into Hooke's law, we get:
Fs = -k * x = -250 N/m * 0.20 m = -50.0 N
Now we can set up the equations of motion for each block.
For Block 1:
Force1 - T = mass1 * acceleration1
For Block 2:
T + Fs - Force2 = mass2 * acceleration2
Since the pulley is light and frictionless, the accelerations of both blocks will be the same in magnitude and direction. We will denote the acceleration as a.
Simplifying the equations, we have:
Force1 - T = mass1 * a
T + Fs - Force2 = mass2 * a
Substituting the known values, we get:
196.0 N - T = 20.0 kg * a
T - 50.0 N - 294.0 N = 30.0 kg * a
Rearranging the equations, we have:
T = 196.0 N - 20.0 kg * a
T = 50.0 N + 294.0 N - 30.0 kg * a
Setting the two expressions for T equal to each other, we can solve for the value of a:
196.0 N - 20.0 kg * a = 50.0 N + 294.0 N - 30.0 kg * a
176.0 N = 10.0 kg * a
a = 176.0 N / 10.0 kg = 17.6 m/s^2
Now, we have the acceleration of both blocks. We can find the final velocity of each block by using the kinematic equation:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
For Block 1, since it starts from rest and moves down an incline 20.0 cm (0.20 m), we have:
v1^2 = 0 + 2 * 17.6 m/s^2 * 0.20 m
v1 = sqrt(0.704) m/s ≈ 0.84 m/s
For Block 2, since it starts from rest and moves up 20.0 cm (0.20 m), we have:
v2^2 = 0 + 2 * 17.6 m/s^2 * 0.20 m
v2 = sqrt(0.704) m/s ≈ 0.84 m/s
Therefore, the speed of Block 1 and Block 2 when Block 2 is 20.0 cm above the floor is approximately 0.84 m/s.