Objects of masses m1 = 4.00 kg and m2 = 9.00 kg are connected by a light string that passes over a frictionless pulley as in Figure P4.48. The object m1 is held at rest on the floor, and m2 rests on a fixed incline of θ = 42.0°. The objects are released from rest, and m2 slides 1.10 m down the incline in 3.60 s.
(a) Determine the acceleration of each object.
m1 in m/s2 upward
m2 in m/s2 down the incline
(b) Determine the tension in the string.
in N
(c) Determine the coefficient of kinetic friction between m2 and the incline.
To solve this problem, we can use Newton's second law of motion (F=ma) and apply it to each object separately. Let's break it down into parts.
(a) To determine the acceleration of each object, we need to find the net force acting on them.
For mass m1:
The only force acting on mass m1 is the tension in the string pulling it upward. We can use Newton's second law to find the acceleration:
T - m1g = m1a₁ --(1)
where T represents the tension, m1 represents the mass, g is the acceleration due to gravity, and a₁ is the acceleration of mass m1.
For mass m2:
The forces acting on mass m2 are the gravitational force downward and the tension in the string acting upward. We need to break down the gravitational force into two components, one parallel and one perpendicular to the inclined plane. Considering the inclined plane, we have:
mg*sin(θ) - T = m2a₂ --(2)
where θ is the angle of the incline, m2 represents the mass, a₂ is the acceleration of mass m2, and T is the tension.
Since the masses are connected by a string, they have the same acceleration. Therefore, a₁ = a₂ = a.
(b) To find the tension in the string, we can substitute the value of acceleration from equation (1) into equation (2) and solve for T:
m1a + mg*sin(θ) - T = m2a
T = m1a + m2a - mg*sin(θ)
T = (m1 + m2)a - mg*sin(θ)
(c) To determine the coefficient of kinetic friction between m2 and the incline, we need to use the equation for the frictional force:
frictional force = coefficient of friction * normal force
Since m2 is sliding down the inclined plane, the frictional force is acting opposite to the direction of motion. The normal force is given by:
normal force = m2 * g * cos(θ)
The frictional force can be expressed as:
frictional force = coefficient of kinetic friction * normal force
m2 * g * cos(θ) * μk = m2 * a
To find the coefficient of kinetic friction (μk), we can cancel out m2:
g * cos(θ) * μk = a
Now, we have all the equations we need to solve the problem. Substitute the given values for masses, distances, and angles, and solve for the unknowns.
Please provide the values of g, θ, m1, m2, distance, and time given in the problem, and I can help you with the calculations.
To solve this problem, we'll break it down into multiple steps:
Step 1: Determine the acceleration of each object
To find the acceleration of each object, we'll use the following equations:
For object m1:
T - m1 * g = m1 * a1
For object m2:
m2 * g - T = m2 * a2
Where T is the tension in the string, g is the acceleration due to gravity (9.8 m/s^2), a1 is the acceleration of m1 (upward), and a2 is the acceleration of m2 (down the incline).
Step 2: Determine the tension in the string
The tension in the string is the same for both objects. So we can use the equation T = m2 * a2 + m2 * g.
Step 3: Determine the coefficient of kinetic friction between m2 and the incline
To find the coefficient of kinetic friction, we'll use the equation:
μk = (m2 * g - T) / (m2 * g * cosθ)
Where μk is the coefficient of kinetic friction and θ is the angle of the incline (42.0°).
Let's plug in the given values and solve for each part of the problem:
(a) Determine the acceleration of each object:
Using the equations mentioned in Step 1:
For object m1:
T - m1 * g = m1 * a1
For object m2:
m2 * g - T = m2 * a2
Given:
m1 = 4.00 kg
m2 = 9.00 kg
θ = 42.0°
Calculating:
Using T = m2 * a2 + m2 * g (equation from Step 2), we can rewrite the equation for object m1 as follows:
m2 * a2 + m2 * g - m1 * g = m1 * a1
Substituting the known values:
9.00 kg * a2 + 9.00 kg * 9.8 m/s^2 - 4.00 kg * 9.8 m/s^2 = 4.00 kg * a1
Simplifying:
9.00 kg * a2 + 9.00 kg * 9.8 m/s^2 - 4.00 kg * 9.8 m/s^2 = 4.00 kg * a1
81.2 N - 39.2 N = 4.00 kg * a1 (since m2 * g = 9.00 kg * 9.8 m/s^2 = 88.2 N)
a1 = 42.0 N / 4.00 kg
a1 = 10.5 m/s^2 (upward)
Now, to find a2, we'll rearrange the equation for object m2 and solve for a2:
m2 * g - T = m2 * a2
9.00 kg * 9.8 m/s^2 - T = 9.00 kg * a2
88.2 N - T = 9.00 kg * a2
Substituting T = m2 * a2 + m2 * g:
88.2 N - (9.00 kg * a2 + 9.00 kg * 9.8 m/s^2) = 9.00 kg * a2
88.2 N - 88.2 N - 9.00 kg * 9.8 m/s^2 = 9.00 kg * a2
- 9.00 kg * 9.8 m/s^2 = 9.00 kg * a2
a2 = (- 9.00 kg * 9.8 m/s^2) / 9.00 kg
a2 = - 9.8 m/s^2 (down the incline)
(b) Determine the tension in the string:
We can use the equation T = m2 * a2 + m2 * g:
T = 9.00 kg * (- 9.8 m/s^2) + 9.00 kg * 9.8 m/s^2
T = 0 N
So, the tension in the string is 0 N. This means m2 is sliding down the incline due to gravity alone, and m1 is being lifted by m2.
(c) Determine the coefficient of kinetic friction between m2 and the incline:
To find the coefficient of kinetic friction (μk), we'll use the equation:
μk = (m2 * g - T) / (m2 * g * cosθ)
Substituting the known values:
μk = (9.00 kg * 9.8 m/s^2 - 0 N) / (9.00 kg * 9.8 m/s^2 * cos(42.0°))
μk = 88.2 N / (88.2 N * cos(42.0°))
μk = 1.00 / cos(42.0°)
μk ≈ 1.303
Therefore, the coefficient of kinetic friction between m2 and the incline is approximately 1.303.
distancetraveled= 1/2 acceleartion*time^2
solve for acceleration.