A 3.4 kg block is pushed along a horizontal floor by a force of magnitude 21 N at a downward angle θ = 40°. The coefficient of kinetic friction between the block and the floor is 0.26. What the magnitude of the block's acceleration?
A warehouse worker exerts a constant horizontal force of magnitude 89.0 N on a 43.0 kg box that is initially at rest on the horizontal floor of the warehouse. When the box has moved a distance of 1.60 m, its speed is 1.10 m/s. What is the coefficient of kinetic friction between the box and the floor?
Does anyone ever answer these?
To find the magnitude of the block's acceleration, we need to analyze the forces acting on the block.
Let's break down the forces in the x and y directions. In the x-direction, we have the force of magnitude 21 N pushing the block and the friction force opposing the motion. In the y-direction, we have the normal force and the force of gravity.
1. Find the force of friction: The force of friction can be calculated using the equation F_friction = μ * N, where μ is the coefficient of kinetic friction and N is the normal force. The normal force is equal to the weight of the block, which is given by F_weight = m * g, where m is the mass of the block and g is the acceleration due to gravity (approximated as 9.8 m/s^2). Therefore, N = m * g. Substituting these values, we have F_friction = μ * m * g.
2. Resolve the force pushing the block: The force of magnitude 21 N is acting at a downward angle of θ = 40°. To find the force component in the x-direction, we calculate F_push_x = F_push * cos(θ). Similarly, the force component in the y-direction is given by F_push_y = F_push * sin(θ).
3. Calculate the net force in the x-direction: The net force in the x-direction, F_net_x, can be found by subtracting the force of kinetic friction from the force pushing the block (assuming the block is already in motion). Thus, F_net_x = F_push_x - F_friction.
4. Apply Newton's second law: Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the net force in the x-direction is equal to the mass of the block multiplied by its acceleration. Thus, F_net_x = m * a.
Now, we can combine these equations to find the magnitude of the block's acceleration.
F_push_x - F_friction = m * a
F_push * cos(θ) - μ * m * g = m * a
Substituting the given values:
21 N * cos(40°) - 0.26 * 3.4 kg * 9.8 m/s^2 = 3.4 kg * a
Simplifying and solving for a:
a = (21 N * cos(40°) - 0.26 * 3.4 kg * 9.8 m/s^2) / 3.4 kg
a ≈ 1.94 m/s^2
Therefore, the magnitude of the block's acceleration is approximately 1.94 m/s^2.