SOLUTION: A 0.5 kg block initially at rest on a frictionless horizontal surface is acted upon by a force of 8.0 N for a distance of 4.0 m how much kinetic energy does the block gain

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Question 1182591: A 0.5 kg block initially at rest on a frictionless horizontal surface is acted
upon by a force of 8.0 N for a distance of 4.0 m how much kinetic energy does the block gain

Found 4 solutions by mananth, ikleyn, CPhill, n2:
Answer by mananth(16949) About Me  (Show Source):
You can put this solution on YOUR website!


Change in Kinetic energy = Work done on object
Work done = F * D
=8.8N *4m
= 35.2 J
Therefore, block gains
35.2 Joule of Kinetic energy.

Answer by ikleyn(53618) About Me  (Show Source):
You can put this solution on YOUR website!
.
A 0.5 kg block initially at rest on a frictionless horizontal surface is acted
upon by a force of 8.0 N for a distance of 4.0 m how much kinetic energy does the block gain
~~~~~~~~~~~~~~~~~~~~~~


        The solution by @mananth and his answer both are incorrect.
        I came to bring a correct solution.


Change in Kinetic energy = Work done on object

Work done = F * D = 8.0N *4 = 32 J

Therefore, block gains 32 Joule of Kinetic energy.

Solved correctly.

----------------------------

The given mass is irrelevant to the solution.



Answer by CPhill(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Change in Kinetic energy = Work done on object
Work done = F * D
=8.8N *4m
= 35.2 J
Therefore, block gains
35.2 Joule of Kinetic energy.

Answer by n2(54) About Me  (Show Source):
You can put this solution on YOUR website!
.

Yesterday (Jan.19, 2026) I witnessed strange behavior from @CPhill on this forum.

A couple of days ago I refuted an incorrect solution provided by @mananth at this spot.
@mananth's solution was incorrect because he used incorrect numbers for his calculations.

As a result, @mananth's solution had nothing in common with the correct solution - which is why
I redid/corrected/fixed it.

Now @CPhill has copied and reposted this incorrect solution by @mananth again.

This is not the only such action by @CPhill.

Yesterday, @CPhill made several (about 15) other similar actions of the same kind with other posts
where I refuted @mananth's solutions.

I consider these @CPhyll's actions to be wrong, leading to a distortion of the truth on this forum.
Therefore, I strongly protest against such actions by @CPhill and consider it necessary that visitors
to this forum be aware of this.


I recommend to a reader to ignore the post by @CPhill.