How Much Work Does It Take to Fire a 2kg Projectile at 50m/s?

AI Thread Summary
To determine the work done in firing a 2.0-kg projectile at 50 m/s, the work-energy theorem is applied, using the equation W = delta K, where K is kinetic energy calculated as K = 1/2mv^2. By substituting the values, K is found to be 2500 J, indicating that this is the total work required. The simplicity of this calculation highlights the effectiveness of the work-energy theorem compared to other methods that involve force or distance. Some participants noted that alternative approaches could complicate the solution unnecessarily. Ultimately, the discussion reinforces the efficiency of using the work-energy theorem for such problems.
mmiller9913
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Homework Statement



Determine the amount of work done in firing a 2.0-kg projectile with an initial speed of 50 m/s. Neglect any effects due to air resistance

Homework Equations



W = delta K
K = 1/2mv^2

The Attempt at a Solution



just take K= 1/2(2)(50)^2 to get 2500...it seems to easy?
 
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mmiller9913 said:

Homework Statement



Determine the amount of work done in firing a 2.0-kg projectile with an initial speed of 50 m/s. Neglect any effects due to air resistance

Homework Equations



W = delta K
K = 1/2mv^2

The Attempt at a Solution



just take K= 1/2(2)(50)^2 to get 2500 J[/color]...it seems to easy?
It only seems easy because you were able to astutely note that the work-energy theorem would give you the answer for total work done. Others might have tried to find force/distance/acceleration etc., to use in calculating the work done, but to no avail.
 

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