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Rama09 [41]
4 years ago
13

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
Physics
1 answer:
mel-nik [20]4 years ago
5 0

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

<u>Explanation:</u>

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.

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