A block of mass m1 = 27 kg rests on a wedge of angle θ = 54∘ which is itself attached to a table (the wedge does not move in this problem). An inextensible string is attached to m1, passes over a frictionless pulley at the top of the wedge, and is then attached to another block of mass m2 = 6 kg. The coefficient of kinetic friction between block 1 and the plane is μ = 0.8. The string and wedge are long enough to ensure neither block hits the pulley or the table in this problem, and you may assume that block 1 never reaches the table. take g to be 9.81 m/s2.
The system is released from rest as shown above, at t = 0.
(a) Find the magnitude of the acceleration of block 1 when it is released (in m/s2).
(b) How many cm down the plane will block 1 have traveled when 0.95 s has elapsed?
m₁a=m₁gsinα-T-F(fr)… (1)
0=N-m₁gcosα………….(2)
m₂a=T-m₂g…………….(3)
F(fr)=μN=μm₁gcosα
sum of (1) and (3):
a(m₁+m₂)=m₁gsinα-T-F(fr)+ T-m₂g=
= m₁gsinα - μm₁gcosα - m₂g
a=g(m₁sinα - μm₁cosα - m₂)/(m₁+m₂) =…
s=at₂/2=….
To find the magnitude of the acceleration of block 1 when it is released, we can use Newton's second law.
(a) The first step is to draw the free-body diagrams for both blocks.
For block 1:
- There is a gravitational force (mg) acting straight down.
- There is a normal force (N) acting perpendicular to the incline.
- There is a frictional force (f) acting parallel to the incline and opposing the motion.
- There is also tension in the string (T) acting upward.
For block 2:
- There is a gravitational force (m2 * g) acting straight down.
- There is tension in the string (T) acting upward.
Now, let's analyze the forces acting on block 1 parallel to the incline:
- The component of the gravitational force acting parallel to the incline is mg * sin(θ).
- The frictional force is μ * N, where N is the normal force.
- The tension in the string is T.
The net force on block 1 in the direction of motion is given by:
F_net = T - mg * sin(θ) - μ * N
Next, let's analyze the forces acting on block 1 perpendicular to the incline:
- The component of the gravitational force acting perpendicular to the incline is mg * cos(θ).
- The normal force is equal to the perpendicular component of the gravitational force, which is mg * cos(θ).
From the components of the forces, we can determine the normal force:
N = mg * cos(θ)
Now, we can substitute the expression for N into the equation for the net force:
F_net = T - mg * sin(θ) - μ * mg * cos(θ)
Since the system is released from rest, the tension in the string is equal to the weight of block 2:
T = m2 * g
Substituting this value back into the equation for the net force:
F_net = m2 * g - mg * sin(θ) - μ * mg * cos(θ)
Now, the net force can be related to the acceleration using Newton's second law:
F_net = m1 * a
Substituting the expression for the net force:
m1 * a = m2 * g - mg * sin(θ) - μ * mg * cos(θ)
Simplifying and solving for a:
a = (m2 * g - mg * sin(θ) - μ * mg * cos(θ)) / m1
(a) Plugging in the given values:
a = (6 kg * 9.81 m/s^2 - 27 kg * 9.81 m/s^2 * sin(54∘) - 27 kg * 9.81 m/s^2 * 0.8 * cos(54∘)) / 27 kg
(a) Calculating the above expression will give you the magnitude of the acceleration of block 1 when it is released.
Now, let's move on to the second part of the question.
To find how many centimeters down the plane block 1 will have traveled when 0.95 seconds have elapsed, we can use the equations of motion for constant acceleration.
(b) The distance traveled by block 1 can be calculated using the following equation:
d = ut + (1/2) * a * t^2
In this case, the initial velocity (u) is 0 m/s since the system is released from rest.
Plugging in the given values:
d = 0 * 0.95 + (1/2) * a * (0.95)^2
(b) Solving the above expression will give you the distance traveled by block 1 in centimeters when 0.95 seconds have elapsed.