A 0.5 kg object, initially at rest, is pulled to the right along a frictionless horizontal surface by a constant horizontal force of 25 N for adistance of 0.5m.
a. What is the work done by the force?
b. What is the change in the kinetic energy of the block?
c. What is the speed of the block after the force is removed?
Help asap plz

Answers

Answer 1
Answer:

Answer:

A.) 12.5 J

B.) 12.5 J

C.) 7.1 m/s

Explanation:

Given that a 0.5 kg object, initially at rest, is pulled to the right along a frictionless horizontal surface by a constant horizontal force of 25 N for a distance of 0.5m.

a. What is the work done by the force?

Work done = force × distance

Work done = 25 × 0.5

Work done = 12.5 J

b. What is the change in the kinetic energy of the block?

Work done = energy

Change in Kinetic energy = work done

Change in kinetic energy = 12.5 J

c. What is the speed of the block after the force is removed?

Kinetic energy = 1/2mV^2

12.5 = 1/2 × 0.5 × V^2

25 = 0.5V^2

V^2 = 25/0.5

V^2 = 50

V = 7.1 m/s


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Answers

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Answers

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Answers

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Answers

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