A block of mass of 24.0 kg is sitting on a frictionless surface. If the block is initially at rest, and a force of 3.0 N parallel to the surface is applied for 5.0 seconds, what is the final speed of the block?.
The acceleration is given by
a = F/m = 1/8 m/s^2
The distance it moves in t = 5.0 s is
X = (1/2)*a*t^2
= (1/16)*25 = 1.563 m
To find the final speed of the block, we can use Newton's second law, which states that force (F) is equal to mass (m) multiplied by acceleration (a), or F = ma.
In this case, the force applied to the block is 3.0 N and the mass of the block is 24.0 kg. So, we can rearrange the formula to solve for acceleration:
a = F/m
Plugging in the values, we have:
a = 3.0 N / 24.0 kg
a = 0.125 m/s^2
Now, we can use the kinematic equation that relates acceleration, initial velocity, final velocity, and time:
vf = vi + at
Since the block is initially at rest (vi = 0), we can simplify the equation to:
vf = at
Plugging in the values of acceleration (a = 0.125 m/s^2) and time (t = 5.0 s), we get:
vf = 0.125 m/s^2 * 5.0 s
vf = 0.625 m/s
Therefore, the final speed of the block is 0.625 m/s.