A 9 kg block initially at rest is pulled to the
right along a horizontal, frictionless surface
by a constant, horizontal force of 18.8 N.
Find the speed of the block after it has
moved 3.3 m.
Answer in units of m/s.
Net force= ma
find a.
vf^2=2a*d=2Fd/m
F=MA>
sqrt(2AX)
distance is X
To find the speed of the block after it has moved 3.3 m, we can use Newton's second law of motion. This law states that the acceleration of an object is equal to the net force acting on it divided by its mass.
First, let's calculate the acceleration of the block. We can use the formula:
acceleration = net force / mass
The net force acting on the block is the force applied to it, which is 18.8 N. The mass of the block is 9 kg. Plugging these values into the formula, we get:
acceleration = 18.8 N / 9 kg
Now, we can use the kinematic equation to find the final velocity of the block after it has moved 3.3 m. The equation we can use is:
vf^2 = vi^2 + 2 × acceleration × distance
In this case, the initial velocity is 0 m/s since the block starts at rest. The acceleration is calculated previously, and the distance is given as 3.3 m. Plugging in these values into the formula, we get:
vf^2 = 0 + 2 × (acceleration) × (distance)
Now, we can solve for vf by taking the square root of both sides of the equation:
vf = √(2 × acceleration × distance)
Plugging in the values for acceleration and distance, we get:
vf = √(2 × (18.8 N / 9 kg) × 3.3 m)
Now, we can calculate the value for vf:
vf ≈ √(2 × (18.8 N / 9 kg) × 3.3 m)
≈ √(2 × 2.088 N × 3.3 m)
≈ √(13.76608 N·m)
≈ 3.71 m/s
Therefore, the speed of the block after it has moved 3.3 m is approximately 3.71 m/s.