Math  /  Algebra

Question1) A 10.31kg10.31-\mathrm{kg} box is at rest when Joseph begins pushing it with a force of +18.66 N , causing it to move 3.64 m . Neglect friction. a) How much work did Joseph do on the box? \begin{tabular}{|l|l|} \hline & J \\ \hline \end{tabular} b) How fast is the box moving after 3.64 m ? \begin{tabular}{|ll|} \hline 3.63 m/s3.63 \mathrm{~m} / \mathrm{s} \\ \hline \end{tabular}

Studdy Solution
Substitute the values and calculate:
v=2×67.9704J10.31kg v = \sqrt{\frac{2 \times 67.9704 \, \text{J}}{10.31 \, \text{kg}}}
v=135.940810.31 v = \sqrt{\frac{135.9408}{10.31}}
v=13.186 v = \sqrt{13.186}
v3.63m/s v \approx 3.63 \, \text{m/s}
The box is moving at 3.63m/s \boxed{3.63} \, \text{m/s} after 3.64 m.

View Full Solution - Free
Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord