Final answer:
The force exerted on the block by the arm is 3.2 N. The relative acceleration of the block with respect to the arm is 0.628 m/s².
Step-by-step explanation:
To determine the force exerted on the block by the arm, we can use Newton's second law for rotation. The torque exerted by the arm is equal to the moment of inertia times the angular acceleration. In this case, the moment of inertia of the block B can be calculated as m*r^2, where m is the mass of the block B and r is the distance from the rotation axis to the block. So, the torque is given by: T = I * α
Using the given values, where m = 0.5 kg, r = 0.8 m, and α = 10 rad/s², we have:
T = (0.5 kg * (0.8 m)^2) * 10 rad/s² = 3.2 N*m
Therefore, the force exerted on the block by the arm is 3.2 N.
To determine the relative acceleration of the block with respect to the arm, we can use the equation for centripetal acceleration. The centripetal acceleration is given by: a = r * ω²
Where r is the distance from the rotation axis to the block and ω is the angular velocity. Since the velocity of the block is zero, we can use the equation ω² = α * θ, where θ is the angle. Plugging in the given values of r = 0.8 m, α = 10 rad/s², and θ = 45°, we have:
ω² = 10 rad/s² * (45° * π/180°) = 0.785 rad²/s²
Then the relative acceleration is:
a = 0.8 m * 0.785 rad²/s² = 0.628 m/s²
Therefore, the relative acceleration of the block with respect to the arm is 0.628 m/s².