Physics · Work done by a constant force and a variable force, kinetic and potential energies, work-energy theorem, power
object has initial velocity \boldsymbol{m} / \boldsymbol{s} . \) The total work
\( \mathbf{A} \) \( 3 k g \) object has initial velocity \( (6 \hat{i}- \) \( \mathbf{2} \hat{\boldsymbol{j}}) \boldsymbol{m} / \boldsymbol{s} . \) The total work done on the object if its velocity changes to \( (8 \hat{i}+ \) \( 4 \hat{j}) m / s \) is :
- A. 2165 J 52665.53
- B. \( 44 J \)
- C. \( 60 J \)
- D. \( 120 J \)
Step-by-step solution
Work done equals change in kinetic energy. Initial KE = 0.5 * 3 kg * (6^2 + (-2)^2) = 60 J. Final KE = 0.5 * 3 kg * (8^2 + 4^2) = 120 J. Change = 60 J.
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