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19 votes
How much work
was done to
accelerate a 2.2
kg object from
rest to 26m/s?

1 Answer

6 votes

Answer:

Step-by-step explanation:

work done = change in kinetic energy, assuming there are no energy losses.

work done = 1/2 × m × change in v^2

= 1/2 × 2.2 × ( 26^2 - 0^2)

= 1/2 × 2.2 × 676

= 743.6 joules

User Suneesh Ambatt
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