Question

If a 0.50-kg block initially at rest on a frictionless, horizontal surface is acted upon by a force of 3.7 N for a distance of 6.8 m, then what would be the block's velocity? m/s

Answer 1

The block's velocity is determined as 10.03 m/s.

Explanation:

According to work energy theorem, the work done on an object is equal to the change in kinetic energy of the object.

So, work done = Kinetic energy

$\text {Force } \times \text { displacement }=\frac{1}{2} \times m \times v^{2}$

Thus, the velocity can be determined as

$\text {velocity}=\sqrt{\frac{2 \times \text {Force} \times \text {displacement}}{m}}$

$\text {velocity}=\sqrt{\frac{2 \times 3.7 \times 6.8}{0.50}}=\sqrt{\frac{50.32}{0.50}}=\sqrt{100.64}$

Velocity = 10.03 m/s.

So the block's velocity is determined as 10.03 m/s.