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.

Step-by-step 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 } * \text { displacement }=(1)/(2) * m * v^(2)`

Thus, the velocity can be determined as

`\text {velocity}=\sqrt{\frac{2 * \text {Force} * \text {displacement}}{m}}`

`\text {velocity}=\sqrt{(2 * 3.7 * 6.8)/(0.50)}=\sqrt{(50.32)/(0.50)}=√(100.64)`

Velocity = 10.03 m/s.

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