A 2.60 kg box with an initial speed of 4.50 m/s at the bottom of a 30.0degree in

Question

A 2.60 kg box with an initial speed of 4.50 m/s at the bottom of a 30.0degree incline is pushed up the incline by a 15.0 N horizontal force. The coefficient of friction between the box and the incline is 0.250. What is the magnitude of the frictional force that opposes the motion? What is the maximum distance the box travels up the incline? Use energy considerations. Use Newton's second law to calculate the acceleration of the box up the incline. Use your answer from (c) and calculate the maximum distance the box travels up the incline using kinematic equations.

Explanation / Answer

a) well , Fr = k x (vertical force)

so Fr = 0.25 (F sin30 + W cos30)

=> Fr = 0.25 (7.5 + 2.6(9.8)(0.866) )

=> Fr = 7.4 N <<<<<<<<<<

b) at the maximum distance v= 0 , hence KE = 0

assuming the distance = x , then the max height is :

h = x sin30 = x/2

applying conservation of energy law u get :

KE1 + PE1 + F.x = KE2 + PE2 + Fr.x

hence 1/2 (2.6) (4.5)2 + 0 + 15x = 0 + 2.6(9.8) (x/2) + 7.4 x

so 26.325 + (15cos30)x = 12.74x + 7.4x

Hence x = 26.325 / 7.14 3.7 m <<<<<<

c) Fcos30 - Fr - Wsin30 = ma

=> 13 - 7.4 - 12.74 = 2.6 a

so a = - 2.746 m/s2 <<<<<<<<<<<

d) use V2 = Vo2 + 2ax

so 0 = (4.5)^2 - 5.49 x

so x 3.7 m <<<<<<<<<

same answer for (b) and (d) shows that the calculations are correct

Hope it's clear.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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