1- A mass m is pushed against a spring of spring constant k = 12000 N/m. The mass slides on a horizontal smooth surface except for a rough one meter long section, and then it encounters a loop of radius R as shown in the figure. The mass just barely makes it around the loop without losing contact with the track (i.e. the normal force at the top of the loop is zero). The coefficient of kinetic friction (k) between the mass and the rough surface is 0.35, m = 2 kg and R = 1 m.
a) Calculate the speed of the mass at the top of the loop. b) Calculate the energy of the mass lost due to friction. c) Calculate the distance the spring was compressed.
To solve this problem, we can break it down into three parts: (a) finding the speed of the mass at the top of the loop, (b) calculating the energy lost due to friction, and (c) determining the distance the spring was compressed.
(a) Calculating the speed of the mass at the top of the loop:
At the top of the loop, the normal force is zero. This means that the gravitational force and the centrifugal force provide the only vertical forces. Equating these forces, we have:
mg = mv^2 / R
Where:
m = mass of the object (2 kg)
g = acceleration due to gravity (9.8 m/s^2)
v = velocity at the top of the loop
Rearranging the equation, we can solve for v:
v^2 = gR
v = √(gR)
Substituting the given values:
v = √(9.8 * 1)
v = √9.8
v ≈ 3.13 m/s
Therefore, the speed of the mass at the top of the loop is approximately 3.13 m/s.
(b) Calculating the energy lost due to friction:
The energy lost due to friction can be calculated using the work-energy principle. The work done by friction is given by:
W = μk * mg * d
Where:
μk = coefficient of kinetic friction (0.35)
m = mass of the object (2 kg)
g = acceleration due to gravity (9.8 m/s^2)
d = distance traveled on the rough surface (1 m)
Substituting the given values:
W = 0.35 * 2 * 9.8 * 1
W ≈ 6.86 J
Therefore, the energy lost due to friction is approximately 6.86 Joules.
(c) Calculating the distance the spring was compressed:
The energy lost due to friction is equal to the potential energy stored in the spring when it was compressed. The potential energy of a spring is given by:
PE = (1/2) * k * x^2
Where:
k = spring constant (12000 N/m)
x = distance the spring was compressed
Substituting the given value:
6.86 J = (1/2) * 12000 * x^2
Simplifying the equation:
x^2 = (6.86 J * 2) / 12000
x^2 ≈ 0.001143
Taking the square root of both sides:
x ≈ √0.001143
x ≈ 0.034 m
Therefore, the distance the spring was compressed is approximately 0.034 meters.
To solve this problem, we will need to consider the conservation of mechanical energy and the forces acting on the mass.
a) Calculating the speed of the mass at the top of the loop:
The mass at the top of the loop just barely makes it around the loop without losing contact with the track. This means that the normal force at the top of the loop should be equal to zero.
At the top of the loop, the only forces acting on the mass are gravity (mg) and the normal force (N). Since the normal force is zero, the net force at the top of the loop is mg.
The net force is given by:
F_net = m * a
For circular motion, the centripetal force is given by:
F_centripetal = m * v^2 / r
Since the net force and the centripetal force are equal at the top of the loop, we can equate them:
m * g = m * v^2 / r
Rearranging the equation to solve for v:
v^2 = g * r
We have g = 9.8 m/s^2 and r = 1 m, so:
v^2 = 9.8 * 1
v = √9.8
v ≈ 3.13 m/s
Therefore, the speed of the mass at the top of the loop is approximately 3.13 m/s.
b) Calculating the energy lost due to friction:
The energy lost due to friction can be calculated using the work-energy principle.
The work done by friction is given by:
W_friction = μk * N * displacement
In this case, the displacement is the length of the rough section of the track, which is 1 meter. We need to find the normal force to determine the work done by friction.
Since the normal force at the top of the loop is zero, the only vertical force acting on the mass is its weight (mg). The normal force is zero when the vertical component of the weight is equal to the centripetal force acting on the mass.
The vertical component of the weight is m * g * cos(theta), where theta is the angle between the vertical direction and the track. At the top of the loop, cos(theta) = 1, so the normal force is equal to mg.
Therefore, the work done by friction is:
W_friction = μk * mg * displacement
Substituting the given values:
W_friction = 0.35 * 2 * 9.8 * 1
W_friction ≈ 6.86 J
Therefore, the energy lost due to friction is approximately 6.86 Joules.
c) Calculating the distance the spring was compressed:
To calculate the distance the spring was compressed, we can use Hooke's law.
Hooke's law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
The force exerted by the spring is given by:
F_spring = -kx
Where k is the spring constant and x is the displacement of the spring.
Since the mass is just barely making it around the loop, the maximum displacement of the spring occurs at the bottom of the loop when the normal force is maximum.
At the bottom of the loop, the force exerted by the spring is equal to the sum of the weight (mg) and the centripetal force (m * v^2 / r).
Therefore, we can equate the two forces:
-kx = mg + m * v^2 / r
Solving for x, the displacement of the spring:
x = -(mg + m * v^2 / r) / k
Substituting the given values:
x = -((2 * 9.8) + 2 * (3.13)^2 / 1) / 12000
x ≈ -0.097 m
Since displacement cannot be negative, the spring was compressed by approximately 0.097 meters.
Therefore, the distance the spring was compressed is approximately 0.097 meters.