A block of mass 0.22 kg can slide on a frictionless incline (Ø = 30º) and is connected via a string to a vertical spring (k = 3.6 N/m). The system starts at rest with the spring at its natural length. (a) What is the maximum distance along the incline through which the block slides? (b) What is the block's speed when it has slid 0.4 m along the slope?
To solve this problem, we will use energy conservation principles.
(a) To find the maximum distance along the incline through which the block slides, we need to determine the point where the potential energy of the spring converts into kinetic energy of the block.
The potential energy stored in the spring when it is stretched or compressed by a distance 'x' can be given by:
U_spring = (1/2) k x^2
Where 'k' is the spring constant. In this case, k = 3.6 N/m.
The initial potential energy of the spring is zero because the spring is at its natural length.
When the block slides through a distance 'd' along the incline, it gains potential energy due to its height above the ground. This potential energy can be given by:
U_gravity = m g h
Where 'm' is the mass of the block and 'h' is the vertical height of the block above the ground.
The height 'h' can be calculated using the angle of the incline (Ø) and the distance 'd' as follows:
h = d sin(Ø)
Therefore, the total energy of the system (spring and block) is conserved, so we can equate the potential energy of the spring and the potential energy gained by the block due to its height:
(1/2) k x^2 = m g d sin(Ø)
Now we can solve for 'd':
d = (1/2) k x^2 / (m g sin(Ø))
Substituting the given values:
k = 3.6 N/m
m = 0.22 kg
g = 9.8 m/s^2
Ø = 30º (which is equal to π/6 radians)
d = (1/2) * 3.6 * x^2 / (0.22 * 9.8 * sin(π/6))
Simplifying the expression:
d = (0.818 x^2) / 0.262
So, the maximum distance along the incline through which the block slides is (0.818/0.262) times the square of the displacement of the spring, 'x'.
(b) To find the block's speed when it has slid 0.4 m along the slope, we can use the principle of conservation of mechanical energy.
The initial mechanical energy of the system is zero since the block was at rest.
When the block slides a certain distance 'd', its potential energy converts into kinetic energy. So, we can write:
(1/2) m v^2 = m g d sin(Ø)
Simplifying the expression, we get:
v^2 = 2 g d sin(Ø)
Substituting the given values:
d = 0.4 m
g = 9.8 m/s^2
Ø = 30º (which is equal to π/6 radians)
v^2 = 2 * 9.8 * 0.4 * sin(π/6)
Simplifying further:
v^2 = 3.92 * sin(π/6)
v = √(3.92 * sin(π/6))
So, the block's speed when it has slid 0.4 m along the slope is the square root of 3.92 times the sine of π/6 radians.
To solve this problem, we can use the principles of energy conservation and trigonometry.
(a) To find the maximum distance along the incline through which the block slides, we need to find the point where all the potential energy is converted into kinetic energy.
Let's analyze the forces acting on the block. There is gravity acting vertically downward, which can be resolved into two components: one parallel to the incline and one perpendicular to it.
The component of gravity parallel to the incline (F_parallel) is given by: F_parallel = m * g * sin(Ø)
Where:
m = mass of the block = 0.22 kg
g = acceleration due to gravity = 9.8 m/s^2
Ø = angle of the incline = 30º
The potential energy of the spring (PE_spring) at maximum compression is equal to the work done by the component of gravity parallel to the incline (F_parallel) over the maximum distance (d_max) the block slides.
PE_spring = F_parallel * d_max
Since the spring force (F_spring) is directly proportional to the displacement from its natural length, we can write:
F_spring = -k * d
Where:
k = spring constant = 3.6 N/m
d = displacement from the natural length of the spring
At maximum compression, the displacement is equal to the maximum distance the block slides, which we can denote as x.
F_spring = -k * x
Setting F_spring equal to F_parallel, we get:
-k * x = m * g * sin(Ø)
Solving for x:
x = (m * g * sin(Ø)) / k
Substituting the given values:
x = (0.22 kg * 9.8 m/s^2 * sin(30º)) / 3.6 N/m
x ≈ 0.2386 m
Therefore, the maximum distance along the incline through which the block slides is approximately 0.2386 meters.
(b) To find the block's speed when it has slid 0.4 m along the slope, we need to use the principles of energy conservation.
The initial potential energy of the spring (PE_spring_i) is zero because it starts at its natural length. The initial potential energy (PE_gravity_i) is also zero since the block starts at rest at the bottom of the incline.
At a displacement of 0.4 meters (d = 0.4 m), the potential energy is given by:
PE_spring_f = 0.5 * k * d^2
The final kinetic energy (KE_f) is given by:
KE_f = 0.5 * m * v^2
Applying the principle of energy conservation, we have:
PE_spring_i + PE_gravity_i = PE_spring_f + KE_f
0 + 0 = 0.5 * k * d^2 + 0.5 * m * v^2
Simplifying the equation, we get:
0.5 * m * v^2 = 0.5 * k * d^2
Solving for v:
v = sqrt[(k * d^2) / m]
Substituting the given values:
v = sqrt[(3.6 N/m * (0.4 m)^2) / 0.22 kg]
v ≈ 1.862 m/s
Therefore, the block's speed when it has slid 0.4 m along the slope is approximately 1.862 m/s.