A 20.0 kg crate is pushed across a rough floor for a distance of 90.0 cm. When the force pushing the block is equal to 80.0 N and applied at an angle of 18.0 below horizontal to the block, the block moves at a constant speed of 2.10 m/s.
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To solve this problem, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration. In this case, since the block is moving at a constant speed, we know that its acceleration is zero.
The first step is to resolve the applied force into its horizontal and vertical components, since the force is acting at an angle to the block.
The horizontal component of the applied force is given by F_horizontal = F * cos(theta), where F is the magnitude of the force and theta is the angle with respect to the horizontal. Substituting the given values, we have:
F_horizontal = 80.0 N * cos(18.0 degrees) = 75.898 N
Since the block is moving at a constant speed, the horizontal component of the applied force must be equal in magnitude to the force of friction acting on the block. So we have:
F_horizontal = force of friction
Next, we can use the equation for the force of friction, which is given by the product of the coefficient of friction (μ) and the normal force (N) acting on the block. The normal force can be calculated using the weight of the block, which is equal to the product of its mass (m) and the acceleration due to gravity (g).
Weight of the block, W = m * g
Normal force, N = W
Given that the mass of the block is 20.0 kg and the acceleration due to gravity is approximately 9.8 m/s^2, we have:
W = 20.0 kg * 9.8 m/s^2 = 196 N
N = 196 N
Substituting the values, we have:
F_horizontal = μ * N
Since the block moves at a constant speed, the force of friction is equal to the coefficient of friction times the normal force. Therefore:
μ * N = 75.898 N
Solving for the coefficient of friction (μ), we have:
μ = 75.898 N / 196 N = 0.387
Therefore, the coefficient of friction is approximately 0.387.
Note: The vertical component of the applied force is not used in this calculation since it does not affect the horizontal motion of the block.