In the figure, a block of mass m = 3.09 kg is pulled along a horizontal frictionless floor by a cord that exerts a force of magnitude F = 13.7 N at an angle θ = 23.0°. (a) What is the magnitude of the block's acceleration? (b) The force magnitude F is slowly increased. What is its value just before the block is lifted (completely) off the floor? (c) What is the magnitude of the block's acceleration just before it is lifted (completely) off the floor?

Wb = M*g = 3.09 * 9.8 = 30.3 N.

Fn = 30.3 - 13.7*sin23 = 24.9 N. = Normal.

a. a = 13.7*Cos23/3.09 = 4.08 m/s^2.

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To find the answers to these questions, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = m * a).

(a) To find the magnitude of the block's acceleration, we can first resolve the force acting on the block into its x and y components. The x-component of the force is given by F_x = F * cos(θ), and the y-component of the force is given by F_y = F * sin(θ).

The net force acting on the block in the x-direction is equal to the x-component of the force, F_netx = F_x. Since the floor is frictionless, there is no horizontal force opposing the motion of the block. Therefore, F_netx is also equal to the mass of the block multiplied by its acceleration, F_netx = m * a.

Setting these two equations equal to each other, we have m * a = F * cos(θ).

Substituting the given values, m = 3.09 kg, F = 13.7 N, and θ = 23.0°, we can calculate the magnitude of the block's acceleration.

a = (F * cos(θ)) / m = (13.7 N * cos(23.0°)) / 3.09 kg = 3.8 m/s² (rounded to one decimal place)

Therefore, the magnitude of the block's acceleration is 3.8 m/s².

(b) To find the value of the force F just before the block is lifted off the floor, we need to consider the vertical direction. The net force acting on the block in the y-direction is equal to the y-component of the force, F_nety = F_y. This net force should be equal to the weight of the block, which is given by the equation F_nety = m * g (where g is the acceleration due to gravity, approximately 9.8 m/s²).

Setting these two equations equal to each other, we have m * g = F * sin(θ).

Substituting the given values, m = 3.09 kg, g ≈ 9.8 m/s², and θ = 23.0°, we can calculate the value of F just before the block is lifted off the floor.

F = (m * g) / sin(θ) = (3.09 kg * 9.8 m/s²) / sin(23.0°) = 53.6 N (rounded to two decimal places)

Therefore, the value of the force F just before the block is lifted off the floor is 53.6 N.

(c) Just before the block is lifted off the floor, the net force acting on the block in the horizontal direction is still equal to the x-component of the force, F_netx = F_x. As mentioned earlier, there is no horizontal force opposing the motion of the block, so this net force is also equal to the mass of the block multiplied by its acceleration, F_netx = m * a.

Setting these two equations equal to each other, we have m * a = F * cos(θ).

Substituting the given values, m = 3.09 kg, F = 53.6 N, and θ = 23.0°, we can calculate the magnitude of the block's acceleration just before it is lifted off the floor.

a = (F * cos(θ)) / m = (53.6 N * cos(23.0°)) / 3.09 kg = 9.6 m/s² (rounded to one decimal place)

Therefore, the magnitude of the block's acceleration just before it is lifted off the floor is 9.6 m/s².