A block of mass m1 = 28 kg rests on a wedge of angle θ = 53∘ 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 = 4 kg. The coefficient of kinetic friction between block 1 and the plane is μ = 0.9. 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 0.50 s has elapsed?
I will assume the inclined block wins and slips down the hill
Force up on inclined block = T
force down slope on inclined block = 28 (9.81 )sin 53 - 28 (9.81)(.9)cos53
a = acceleration down slope
28 a = 28 (9.81 )sin 53 - 28 (9.81)(.9)cos53 - T
Force down on hang mass = m g = 4*9.81
force up on hang mass = T
a = acceleration up
4 a = T- 4*9.81
or
28 a = 7 T - 28*9.81
so
28 (9.81 )sin 53 - 28 (9.81)(.9)cos53 - T = 7 T - 28*9.81
or
sin 53 - (.9)cos53 = 8 T/(28*9.81) - 1
or
8 T/(28*9.81) = 1 + sin 53 - (.9)cos53
solve for T
then go back and get a
then get d from
d = (1/2) a t^2
remember to multiply by 100 because it asks for cm, not meters
To find the magnitude of the acceleration of block 1 when it is released, we can analyze the forces acting on it.
1. Determine the gravitational force on block 1:
The gravitational force can be calculated using the formula: F_grav = m1 * g, where m1 is the mass of block 1 and g is the acceleration due to gravity. Substituting the given values: m1 = 28 kg and g = 9.81 m/s^2, we get F_grav = 28 kg * 9.81 m/s^2.
2. Determine the component of the gravitational force parallel to the incline:
Since the wedge is inclined at an angle θ = 53∘, we can find the component of the gravitational force parallel to the incline by multiplying the gravitational force by the sine of the angle: F_parallel = F_grav * sin(θ).
3. Determine the force of kinetic friction:
The force of kinetic friction can be calculated using the formula: F_friction = μ * N, where μ is the coefficient of kinetic friction. The normal force N can be calculated as N = m1 * g * cos(θ), where cos(θ) is the cosine of the angle. Substituting the given values: μ = 0.9, m1 = 28 kg, and g = 9.81 m/s^2, we can calculate N = 28 kg * 9.81 m/s^2 * cos(53∘). Then, we can find F_friction = 0.9 * N.
4. Calculate the net force acting on block 1:
The net force is the difference between the parallel component of the gravitational force and the force of kinetic friction: F_net = F_parallel - F_friction.
5. Use Newton's second law to find the acceleration:
According to Newton's second law, F_net = m1 * a, where a is the acceleration of block 1. Rearranging the equation, we get a = F_net / m1. Substituting the calculated values of F_net and m1, we can find the acceleration of block 1.
Now let's move on to part (b) of the question.
To find how many cm down the plane block 1 will have traveled when 0.50 s has elapsed, we need to consider the motion equations.
1. Calculate the initial velocity of block 1:
Since the system is released from rest, the initial velocity is zero: v0 = 0 m/s.
2. Use the equation of motion:
The equation to calculate the displacement of an object in motion is given by: s = v0 * t + (1/2) * a * t^2, where s is the displacement, v0 is the initial velocity, t is the time, and a is the acceleration. Substitute the known values into the equation: s = 0 * 0.50 + (1/2) * acceleration * (0.50)^2.
By substituting the calculated acceleration from part (a) and evaluating this expression, you can find how many cm down the plane block 1 will have traveled after 0.50 s.
Remember to convert the final answer to centimeters if necessary.