Consider the system shown in the figure with = 9.5 and = 15.0 . The angles = 59 and = 32. 1)In the absence of friction, what force would be required to pull the masses at a constant velocity up the fixed inclines? the force is now removed. What is the magnitude of the acceleration of the two blocks?
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To find the force required to pull the masses at a constant velocity up the fixed inclines, you can use the concept of resolving forces into components along the direction of motion.
1) In the absence of friction, the force required to pull the masses at a constant velocity up the fixed inclines can be determined by equating the forces along the direction of motion to the net force in that direction.
Let's consider the vertical component of force (mg) acting on each block:
- For the 9.5 kg block, the component acting upward along the inclined plane is mg(sinθ1), where θ1 is the angle of the inclined plane with the horizontal.
- For the 15.0 kg block, the component acting downward along the inclined plane is mg(sinθ2), where θ2 is the angle of the inclined plane with the horizontal.
Since the system is at a constant velocity, the net force along the direction of motion is zero. Therefore, the force required to overcome the weight of the blocks is given by:
F = mg(sinθ2) - mg(sinθ1)
Substituting the values given in the problem:
F = (9.5 kg)(9.8 m/s^2)(sin32°) - (15.0 kg)(9.8 m/s^2)(sin59°)
Calculate this expression to find the force required to pull the masses at a constant velocity up the fixed inclines.
2) If the force F is now removed, the magnitude of acceleration of the two blocks can be determined using Newton's second law, which states that the net force acting on an object is equal to the product of its mass and acceleration.
Considering the 9.5 kg block:
The net force acting on the block is the component of the weight acting downward along the inclined plane (mg(sinθ1)) minus the frictional force opposing its motion. Since there is no friction mentioned in the problem, the net force is simply mg(sinθ1).
Applying Newton's second law:
mg(sinθ1) = m1 * a
Where m1 is the mass of the 9.5 kg block and a is its acceleration. Solve for a.
Similarly, for the 15.0 kg block:
The net force acting on the block is the component of the weight acting upward along the inclined plane (mg(sinθ2)) plus the force of tension T between the two blocks. Since the force of tension is not mentioned in the problem, you can solve for T from the previous part (where F was calculated) and substitute it into the equation.
Using Newton's second law:
mg(sinθ2) + T = m2 * a
Where m2 is the mass of the 15.0 kg block and a is its acceleration. Solve for a.
Calculate the expressions for a for both blocks to find the magnitude of the acceleration of the two blocks.