This was a two part problem I only need the answer for part 2 bellow This is the

Question

This was a two part problem I only need the answer for part 2 bellow

This is the first part of the problem it has some info, I already have the answer for it = 3.77 m/s^2

The student now starts moving the box up a 14.8 incline, keeping her 160 N force directed at 27 above the line of the incline. If the coefficient of friction is unchanged, what is the new acceleration of the box? Answer in units of m/s2 A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a force of 160 N at an angle of 27 above the horizontal. The box has a mass of 25.3 kg. and the coefficient of friction between box and floor is 0.268. The acceleration of gravity is 9.8 m/s2 . Find the acceleration of the box. Answer in units of m/s2

Explanation / Answer

sum forces perp to the slope

N + F sin theta - m g cos theta = 0

N = 25.3*9.81*cos(14.8 degrees) - 160*sin(27 degrees) = 167.3 N

sum forces parallel to slope

F cos theta - m g sin theta - friction = ma

friction = u N

160*cos(27 degrees) - 25.3*9.81*sin(14.8 degrees) -0.268*167.3 = 25.3*a

a=1.36 m/s^2

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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