43. A block of mass 2.00 kg rests on a ramp tilted at 25.0°. Coefficients of friction for the block/ramp: μs = 0.300, μk = 0.100. A string passing over a high quality pulley connects the block with a bucket as shown below. The bucket contains sand but a hole is punched in the bottom and it starts to run out. When there is a certain amount of sand left in the bucket the block starts to move. (a) What mass of sand is left in the bucket when it starts to move? (b) Find the acceleration rate of the block/bucket system shortly after motion begins.
To find the mass of sand left in the bucket when the block starts to move, we need to consider the forces acting on the block and the maximum static friction force.
(a) The gravitational force acting on the block can be calculated using the mass of the block (2.00 kg) and the acceleration due to gravity (9.8 m/s^2). The gravitational force is given by F_gravity = m * g, where m is the mass and g is the acceleration due to gravity. Therefore, F_gravity = 2.00 kg * 9.8 m/s^2 = 19.6 N.
The maximum static friction force can be calculated using the coefficient of static friction (μs) and the normal force. The normal force is equal to the gravitational force since the block is sitting on the ramp without any vertical acceleration. Therefore, the normal force is 19.6 N.
The maximum static friction force is given by the equation F_max_static_friction = μs * N, where μs is the coefficient of static friction and N is the normal force. Thus, F_max_static_friction = 0.300 * 19.6 N = 5.88 N.
When the block starts to move, the static friction force will reach its maximum value before transitioning to the kinetic friction force. Therefore, the force required to start the motion is equal to the maximum static friction force, which is 5.88 N.
Now, we can consider the forces acting on the block due to the sand in the bucket. The sand in the bucket exerts a force on the block opposite to the direction of the gravitational force. Let's denote the mass of sand in the bucket as 'm_sand'. The force exerted by the sand is given by F_sand = m_sand * g.
When the block starts to move, the force exerted by the sand reaches the value of the maximum static friction force (5.88 N). Therefore, we can set up the equation F_sand = F_max_static_friction:
m_sand * g = 5.88 N.
Rearranging this equation, we can solve for m_sand:
m_sand = 5.88 N / g.
Given that g = 9.8 m/s^2, we can calculate the mass of sand left in the bucket when the block starts to move:
m_sand = 5.88 N / 9.8 m/s^2 ≈ 0.60 kg.
Therefore, the mass of sand left in the bucket when it starts to move is approximately 0.60 kg.
(b) After the motion begins, the friction force transitions from static friction to kinetic friction. The acceleration of the block/bucket system can be found using the net force acting on the block.
The net force in the horizontal direction can be calculated as the difference between the gravitational force acting down the ramp and the kinetic friction force acting up the ramp.
The gravitational force component down the ramp is given by F_gravity_parallel = m * g * sin(θ), where θ is the angle of the ramp (25.0°).
F_gravity_parallel = 2.00 kg * 9.8 m/s^2 * sin(25.0°) ≈ 8.19 N.
The kinetic friction force is given by F_kinetic_friction = μk * N, where μk is the coefficient of kinetic friction and N is the normal force. Since the block is in motion, we use the kinetic friction coefficient (μk = 0.100).
F_kinetic_friction = 0.100 * 19.6 N = 1.96 N.
Now, we can calculate the net force acting on the block:
Net force = F_gravity_parallel - F_kinetic_friction = 8.19 N - 1.96 N = 6.23 N.
To find the acceleration of the block/bucket system, we can use Newton's second law, which states that the net force is equal to the mass of the system multiplied by the acceleration (F_net = m_system * a).
In this case, the system consists of the block and the sand. The mass of the system is the sum of the mass of the block and the mass of the sand left in the bucket.
m_system = m_block + m_sand = 2.00 kg + 0.60 kg = 2.60 kg.
Now we can plug in the known values into Newton's second law to find the acceleration:
6.23 N = 2.60 kg * a.
Solving for a, we find:
a = 6.23 N / 2.60 kg ≈ 2.40 m/s^2.
Therefore, the acceleration rate of the block/bucket system shortly after motion begins is approximately 2.40 m/s^2.
a) 1.39 kg
b) 1.04 m/s^2