Question

If a 0.50 kg block initially at rest on a frictionless, horizontal surface is acted upon by a force of 2.2 N for a distance of 2.1 m, then what would be the block's velocity (in m/s)?

Answer 1

The block's velocity is calculated using the work-energy principle, resulting in a final velocity of 6.08 m/s after a force of 2.2 N acts over a distance of 2.1 m.

Step-by-step explanation:

To find the block's velocity after a force acts on it, we can use the work-energy principle, which states that the work done on an object by a net force will result in a change in the object's kinetic energy. The work (W) done on the block by the force is the product of the force (F) and the distance (d) over which it is applied:

W = F × d

Thus, for a force of 2.2 N acting over a distance of 2.1 m, the work done on the 0.50 kg block is:

W = 2.2 N × 2.1 m = 4.62 J

The change in kinetic energy (ΔKE) is equal to the work done:

ΔKE = W

Since the block is initially at rest, its initial kinetic energy is 0, and the work done on the block is equal to its final kinetic energy:

ΔKE = ½ m v2 - 0 = 4.62 J

Solving for the final velocity (v), we have:

v = √(2 ΔKE / m)

v = √(2 × 4.62 J / 0.50 kg)

v = √(18.48 / 0.50)

v = √(36.96)

v = 6.08 m/s

Therefore, the block's velocity after the force has been applied over the given distance is 6.08 m/s.