A man stacks a small block of wood on top of a larger block and then moves both by pushing the
bottom block. The large block has a mass of 25 kg, and the smaller block has a mass of 15 kg. Thecoefficient of static friction between the two blocks is 0.45. How fast can the man accelerate the blocks without having the smaller block slide off the top?
To find out how fast the man can accelerate the blocks without having the smaller block slide off, we need to calculate the maximum static friction force between the two blocks.
1. Calculate the normal force acting on the smaller block:
The normal force (Fn) is equal to the weight of the block, which is its mass times the acceleration due to gravity (g = 9.8 m/s^2).
Fn = mass * gravity = 15 kg * 9.8 m/s^2 = 147 N
2. Calculate the maximum static friction force (Ff,max):
Ff,max = coefficient of static friction * normal force
Ff,max = 0.45 * 147 N = 66.15 N
3. The maximum static friction force is also equal to the force applied to the blocks by the man (Fapplied) to accelerate them:
Fapplied = Ff,max = 66.15 N
4. Use Newton's second law (F = ma) to find the maximum acceleration (a):
Fapplied = mass * acceleration
66.15 N = (25 kg + 15 kg) * acceleration
66.15 N = 40 kg * acceleration
acceleration = 66.15 N / 40 kg
acceleration ≈ 1.65375 m/s^2
Therefore, the man can accelerate the blocks at a maximum of approximately 1.65375 m/s^2 without having the smaller block slide off the top.
To determine how fast the man can accelerate the blocks without the smaller block sliding off the top, we need to consider the maximum static friction between the two blocks.
The maximum static friction force (F_friction) between two surfaces can be calculated using the equation:
F_friction = μ_s * F_normal
Where:
- μ_s is the coefficient of static friction
- F_normal is the normal force between the two surfaces
First, we need to calculate the normal force (F_normal) acting on the smaller block. The normal force is equal to the weight (mg) of the block, where m is the mass and g is the acceleration due to gravity (9.8 m/s^2).
For the smaller block:
m_smaller = 15 kg
F_normal_smaller = m_smaller * g = 15 kg * 9.8 m/s^2 = 147 N
Next, we can calculate the maximum static friction force (F_friction) between the two blocks using the coefficient of static friction (μ_s). In this case, it's 0.45.
F_friction = μ_s * F_normal_smaller = 0.45 * 147 N = 66.15 N
The maximum static friction force (F_friction) is also equal to the force the man exerts on the larger block (F_man) to accelerate the blocks.
Finally, we can calculate the maximum acceleration (a) the man can achieve without having the smaller block slide off the top using Newton's second law of motion:
F_man = (mass_of_larger_block + mass_of_smaller_block) * a
Rearranging the equation and substituting the values:
66.15 N = (25 kg + 15 kg) * a
66.15 N = 40 kg * a
a = 66.15 N / 40 kg ≈ 1.65 m/s^2
Therefore, the man can accelerate the blocks at a maximum rate of approximately 1.65 m/s^2 without having the smaller block slide off the top.
To determine the maximum acceleration that the man can apply without the smaller block sliding off the top, we need to consider the forces acting on the system.
The force of static friction between the two blocks will provide the necessary force to prevent the smaller block from sliding. The equation for static friction is given by:
fs ≤ μs * N
where fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force. In this case, the normal force on the smaller block is its weight:
N = m * g
where m is the mass of the smaller block and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Now, let's calculate the maximum force of static friction:
fs ≤ 0.45 * m * g
Next, we need to consider the force applied by the man, which is equal to the net force acting on the system:
F = (m1 + m2) * a
where m1 is the mass of the larger block, m2 is the mass of the smaller block, and a is the acceleration of the system.
Since the force applied by the man pushes the blocks to accelerate, it must overcome the force of static friction:
F = fs
Using the equations mentioned above, we can set the force applied by the man equal to the maximum force of static friction:
(m1 + m2) * a = 0.45 * m2 * g
Now, we can solve for the acceleration:
a = (0.45 * m2 * g) / (m1 + m2)
Inserting the given masses and acceleration due to gravity into the equation:
a = (0.45 * 15 kg * 9.8 m/s^2) / (25 kg + 15 kg)
Simplifying:
a ≈ 0.99 m/s^2
Therefore, the man can accelerate the blocks up to approximately 0.99 m/s^2 without having the smaller block slide off the top.