Answer to Question #113812 in Classical Mechanics for Caylin

Question

A) How much work must you do to push a 10kg block of steel across a steel table at a steady speed of 1.0 m/s for 3.0 s?

B) What's the power output while doing so?

Answer 1

As per the given question,

mass of the object $(m)=10kg$

speed of the block $(v)=1.0 m/sec$

time $(t)=3.0 sec$

a) work done = change in kinetic energy $(W)=\dfrac{mv^2}{2}=\dfrac{10\times 1}{2}=5J$

b) Power =rate of change work

$=\dfrac{5}{3} j/sec$