In the following figure m_1 = 20.0kg kg and á alpha = 50.9 ∘ ^\circ. The coefficient of kinetic friction between the block and the incline is ì k = 0.40
What must be the mass m 2 of the hanging block if it is to descend 9.00m m in the first 3.00s s after the system is released from rest?
s=at²/2 +> a=2s/t²=
=2•910⁻³/3²=2•10⁻³ m/s².
m₁a=T-F(fr) -m₁gsinα,
0=m₁gcosα –N,
m₂a=m₂g –T.
F(fr) =μN=μm₁gcosα .
m₁a=T- μm₁gcosα -m₁gsinα,
m₂a=m₂g –T.
m₁a + m₂a= m₂g - μm₁gcosα -m₁gsinα
m₂(g-a) = m₁(a+ μgcosα+gsinα)
m₂=[m₁(a+ μgcosα+gsinα) ]/(g-a)
To determine the mass m2 of the hanging block, we can use the principles of dynamics and kinematics.
Step 1: Calculate the acceleration of the system
Since the system is released from rest, the acceleration can be calculated using the formula:
a = g * (sin(α) - μk * cos(α))
where
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- α is the angle of the incline (50.9°)
- μk is the coefficient of kinetic friction (0.40)
Substituting the given values, let's calculate the value of a:
a = 9.8 * (sin(50.9°) - 0.40 * cos(50.9°))
Step 2: Calculate the final velocity
Since the initial velocity is zero, the final velocity can be calculated using the formula:
vf = vi + a * t
where
- vf is the final velocity
- vi is the initial velocity (zero in this case)
- a is the acceleration calculated in Step 1
- t is the time taken (3.00 s)
Substituting the known values:
vf = 0 + a * 3.00
Step 3: Use kinematic equation to calculate displacement
The displacement of the block can be calculated using the formula:
d = vi * t + 0.5 * a * t^2
where
- d is the displacement
- vi is the initial velocity (zero in this case)
- a is the acceleration calculated in Step 1
- t is the time taken (3.00 s)
Substituting the known values:
9.00 = 0 + 0.5 * a * (3.00)^2
Step 4: Rearrange the equation to solve for a
To solve for the acceleration (a), we can rearrange the equation from Step 3:
a = (2 * d) / (t^2)
Substituting the known values:
a = (2 * 9.00) / (3.00)^2
Step 5: Calculate the mass m2
To determine the mass m2 of the hanging block, we can use Newton's second law:
ΣF = m2 * a
where
- m2 is the mass of the hanging block
- a is the acceleration calculated in Step 4
We can rearrange this equation to solve for m2:
m2 = ΣF / a
Knowing that the force acting on the block is the gravitational force, which can be calculated as:
ΣF = m2 * g
Substituting the known values:
m2 = (m2 * g) / a
Now we can solve for m2.
Keep in mind that it is necessary to approach the hanging rope as massless in this situation.
To solve this problem, we need to analyze the forces acting on the system, which includes the block on the incline and the hanging block.
Let's break the forces into components.
1. For the block on the incline:
- The weight force (mg) is acting vertically downward.
- The normal force (N) is acting perpendicular to the incline.
- The force of kinetic friction (fk) is opposing the motion and acting parallel to the incline.
2. For the hanging block:
- The weight force (mg2) is acting vertically downward.
Now, let's analyze the motion of the system during the first 3.00s:
1. The acceleration of the system can be found using the equation:
a = (m1 * g * sin(alpha) - fk) / (m1 + m2)
where g is the acceleration due to gravity.
Note that the force of kinetic friction (fk) can be calculated using the equation:
fk = µk * N,
where µk is the coefficient of kinetic friction and N is the normal force.
The normal force can be calculated using the equation:
N = m1 * g * cos(alpha).
2. Using the acceleration, we can find the distance covered by the system in 3.00s using the equation:
d = 0.5 * a * t^2,
where t is the time interval.
3. Equating the distance covered to 9.00m, we can solve for the mass of the hanging block (m2).
Now, let's substitute the given values into the equations and solve:
Given:
m1 = 20.0 kg
alpha = 50.9°
µk = 0.40
d = 9.00 m
t = 3.00 s
1. Calculate the normal force:
N = m1 * g * cos(alpha)
= 20.0 kg * 9.8 m/s^2 * cos(50.9°)
2. Calculate the force of kinetic friction:
fk = µk * N
3. Calculate the acceleration:
a = (m1 * g * sin(alpha) - fk) / (m1 + m2)
4. Calculate the distance covered by the system in 3.00 seconds:
d = 0.5 * a * t^2
5. Equate the distance to 9.00m and solve for m2:
9.00m = 0.5 * a * (3.00s)^2
Solving this equation will give you the mass (m2) of the hanging block needed for the system to descend 9.00m in the first 3.00s after being released from rest.