A 60 kg block slides along the top of a 100 kg block. The lighter block has an acceleration of 3.7 m/s2 when a horizontal force F= 350 N is applied. Assuming there is no friction between the bottom 100 kg block and the horizontal frictionless surface but there is friction between the blocks. Find the acceleration of the 100 kg block during the time the 60 kg block remains in contact.
coefficient of friction needed
To find the acceleration of the 100 kg block, we need to consider the forces acting on both blocks and the interaction between them.
1. Let's start by analyzing the forces acting on the 60 kg block:
- Gravitational force (weight): Since there is no friction between the 100 kg block and the surface, the weight of the 60 kg block does not affect its acceleration in this direction.
- Applied force: The horizontal force F = 350 N is the only force acting on the 60 kg block in the positive direction.
- Friction force: There is friction between the blocks since they are in contact. This friction force acts in the opposite direction to the applied force and is responsible for the acceleration of the 60 kg block.
2. Now let's consider the forces acting on the 100 kg block:
- Gravitational force (weight): This force acts vertically downwards.
- Normal force: The 100 kg block exerts a normal force on the 60 kg block due to their contact.
3. From Newton's second law, we know that the net force acting on an object is equal to the product of its mass and acceleration (F = m*a).
4. For the 60 kg block:
- The net force is the applied force minus the friction force: F_net = F_applied - F_friction.
- Using F_net = m*a, we can write the equation as: F_applied - F_friction = m*a.
- Substituting the given values: 350 N - F_friction = 60 kg * 3.7 m/s^2.
5. The friction force between the blocks can be calculated using the coefficient of friction and the normal force:
- Since the surface is frictionless, there is no friction between the 100 kg block and the surface.
- The friction force is given by: F_friction = μ * F_normal, where μ is the coefficient of friction and F_normal is the normal force.
- In this case, F_normal = 100 kg * 9.8 m/s^2 (acceleration due to gravity), as the only vertical force acting on the 100 kg block is its weight.
- The coefficient of friction between the blocks, μ, is not given. To solve for it, we need to consider the limiting friction force (F_friction_max) based on the friction coefficient (μ):
- F_friction_max = μ * F_normal.
- Since the 60 kg block has its maximum acceleration (3.7 m/s^2), the friction force acting on it must also be at its maximum, which is equal to the limiting friction force.
- Therefore, we can write: F_friction = F_friction_max = μ * F_normal.
- Substituting the values in the equation derived in step 4, we get: 350 N - μ * (100 kg * 9.8 m/s^2) = 60 kg * 3.7 m/s^2.
6. Now, solve the equation from step 5 for the coefficient of friction (μ).
7. Once you have calculated the coefficient of friction, substitute it back into the equation from step 4 to solve for the acceleration of the 100 kg block.
Following these steps will allow you to find the acceleration of the 100 kg block during the time the 60 kg block remains in contact.