Chrome File Edit View History Bookmarks People Window Help Hww502091) x C Chegg

Question

Chrome File Edit View History Bookmarks People Window Help Hww502091) x C Chegg Study I Guided Solutio x Tue 8:02 PM c O Need Help? Read 16215997 Previous Answers SerCP10 5.P012. A worker pushing a 35.0-kg wooden crate My Notes o Ask Your Te magnitude at a constant speed for 12.7 m along a wood foor does 370 J of work by applying a constant horizontal force of Fo on the crate. (a) Determine the value of Fo. Your response differs from the correct answer by more than 10%. Double check your calculations. N (b) If the worker now applies a force greater than Fo, describe the subsequent motion of the crate. the worker pushes with greater force the create wil accelerate (it will increase the speed because the worker has done more work. Score: 1 out of 1 Comment (c) Describe what would happen to the crate if the applied force is less than Fo if the gorker pushes with lesser the crate will slow (he does less work and the friction force is the same so the crate wil slow down

Explanation / Answer

(a) The force may be found from the fact that work is:

W = Fx

Solve for F :

F = W / x

= 370J / 12.7m

= 29.13 N

(b) With this force, the crate moves at constant speed. This means that the net forces (F - friction = 0, no acceleration) are zero. If the force applied by the worker is greater, the net forces will be greater than zero, and the crate will accelerate (that is, its speed will increase as long as there is a greater force F applied.

(c) From the discussion in part (b), it can be inferred that a force less than F would mean that the crate would not move. The force of static friction between the crate and the floor would then be larger than F.

Any time a problem says constant speed, it is code for zero acceleration and a balance of forces on the object of interest.

Hope this helps.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.