Consider a horizontal spring of stiffness k = 87 N/m attached
to a 1.5 kg block at one end and a wall at the other. The spring is compressed by an amount x = 20 cm from
its unstretched length, x0.
a. If the mass is released from rest, how much work was done on the mass by the spring as the spring uncompresses and returns to its equilibrium length
b. Suppose the mass loses contact with the spring when the spring is at its equilibrium length. Using the work-kinetic energy theorem, what is the speed of the mass when it loses contact with the spring?
c. The mass slides along the horizontal surface and at some point, it encounters a ramp inclined at an angle of 41 degrees measured with
respect to the horizontal. There is friction between the block and the
ramp with coefficient of friction μ = 0.5. What is the total work done on the block by all outside forces bringing the mass to rest?
d. How far along the ramp, d, does the block slide before coming to rest?
Thanks in advance!
a. To find the work done by the spring on the mass as it uncompresses and returns to its equilibrium length, we can use the formula for the work done by a spring:
Work = (1/2)k(x^2 - x0^2)
where k is the spring stiffness, x is the compression distance, and x0 is the initial compression distance (unstretched length).
Given:
k = 87 N/m
x = 20 cm = 0.2 m
x0 = 0 (since the spring is initially compressed)
Using the values in the formula, we get:
Work = (1/2)(87)(0.2^2 - 0^2)
= (1/2)(87)(0.04)
= 1.74 J
Therefore, the work done by the spring on the mass as it uncompresses and returns to its equilibrium length is 1.74 Joules.
b. To find the speed of the mass when it loses contact with the spring, we can use the work-kinetic energy theorem. According to the theorem:
Work done on an object = Change in kinetic energy of the object
Since the spring loses contact when it returns to its equilibrium length, the work done by the spring is equal to the change in kinetic energy of the mass.
The work done by the spring is already calculated as 1.74 J (from part a).
Since the mass is initially at rest, the initial kinetic energy is 0.
Therefore, the final kinetic energy of the mass is also 1.74 J.
Using the formula for kinetic energy:
Final kinetic energy = (1/2)mv^2
where m is the mass and v is the velocity (which is the same as the speed in this case), we can solve for v:
1.74 J = (1/2)(1.5 kg)v^2
2(1.74 J) / 1.5 kg = v^2
2.32 m/s^2 = v^2
v = √2.32 m/s^2
v ≈ 1.52 m/s
Therefore, the speed of the mass when it loses contact with the spring is approximately 1.52 m/s.
c. To find the total work done on the block by all outside forces bringing the mass to rest, we need to consider the work done against friction and the work done against gravity on the ramp.
The work done against friction can be calculated using the formula:
Work_friction = force_friction * distance
where force_friction = μ * N (friction coefficient * normal force) and distance is the displacement along the ramp.
Given:
μ = 0.5
θ = 41 degrees
m = 1.5 kg
The normal force N can be found using the equation:
N = mg * cos(θ)
where g is the acceleration due to gravity.
N = (1.5 kg)(9.8 m/s^2) * cos(41 degrees)
N ≈ 10.7 N
Now we can calculate the work done against friction:
Work_friction = (0.5)(10.7 N)(d)
Next, we need to calculate the work done against gravity. The component of weight acting along the ramp is given by:
Force_gravity = mg * sin(θ)
Using the same values as before, we find:
Force_gravity = (1.5 kg)(9.8 m/s^2) * sin(41 degrees)
Force_gravity ≈ 9.65 N
The work done against gravity can be calculated as:
Work_gravity = Force_gravity * d
where d is the displacement along the ramp.
Now we can find the total work done on the block by all outside forces:
Total Work = Work_gravity + Work_friction
d. To find how far along the ramp the block slides before coming to rest, we need to determine the point where the total work done on the block is equal to zero.
Using the calculated values for Work_gravity and Work_friction, we can solve the equation:
Total Work = Work_gravity + Work_friction = 0
By substituting the expressions for each term, we get:
(1.5 kg)(9.8 m/s^2) * sin(41 degrees) * d + (0.5)(10.7 N)(d) = 0
Simplifying the equation, we can solve for d:
(9.8 m/s^2) * sin(41 degrees) * d + (0.5)(10.7 N) = 0
Solving for d:
d = -[(0.5)(10.7 N)] / [(9.8 m/s^2) * sin(41 degrees)]
Evaluate the right-hand side to find the value of d.
Please note that the negative sign indicates that the displacement is in the opposite direction of the external forces.