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Question

If a 0.50-kg block initially at rest on a frictionless, horizontal surface is acted upon by a force of 8.9 N for a distance of 2.0 m, then what would be the block's velocity?
Answer:_m/s

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Answer 1 · 2018-02-15T21:42:42

The velocity is #v = 8.44 m/s#.

The work done, w, by that force would be

#w = F*s = 8.9 N*2.0 m = 17.8 J#

Since the friction is zero and the surface is horizontal, all 17.8 J will go toward kinetic energy, KE, of the block.

#"KE" = 17.8 J = 1/2 *m*v^2 = 1/2 *0.5 kg*v^2#

Solving for v,

#v = sqrt((17.8 J*2)/(0.5 kg)) = sqrt(71.2 J/"kg"#

#v = 8.44 m/s#

How did J/kg get changed into m/s?

From the formula for KE, #"KE" = 1/2 *m*v^2#, we see that the Joule is equivalent to #kg*m^2/s^2#.

I hope this helps,
Steve