Two forces, 1 and 2, act on the 5.00 kg block shown in the drawing. The magnitudes of the forces are F1 = 47.5 N and F2 = 34.5 N. What is the horizontal acceleration (magnitude and direction) of the block?
ma=F2-F1•cosα,
a= (F2-F1•cosα)/m
To find the horizontal acceleration of the block, we need to determine the net force acting on the block in the horizontal direction using Newton's second law of motion, which states that the net force on an object is equal to its mass multiplied by its acceleration:
net force = mass * acceleration
In this case, the net force is the vector sum of the two forces, F1 and F2:
net force = F1 + F2
Given that F1 = 47.5 N, F2 = 34.5 N, and the mass of the block is 5.00 kg, we can substitute these values into the equation to find the net force:
net force = 47.5 N + 34.5 N = 82 N
Now, we can rearrange the equation to solve for acceleration:
acceleration = net force / mass
Substituting the values, we get:
acceleration = 82 N / 5.00 kg ≈ 16.4 m/s^2
So, the magnitude of the horizontal acceleration of the block is approximately 16.4 m/s^2.
To determine the direction of the acceleration, we need to consider the forces acting on the block. In this case, F1 is larger than F2, and it is applied in the positive x-direction. Therefore, the block will accelerate in the direction of the larger force, which is the positive x-direction.
Hence, the direction of the acceleration is in the positive x-direction.