Question

Question 4 8 pts A box of mass m = 7 kg is being pushed up a ramp that is angled 15 degrees above horizontal. The pushing force P = 49 N. The coefficient of kinetic friction is 0.215. At a certain time the velocity of the box moving up the ramp is 1.0 m/s. How far will the box been pushed during the next 3.0 seconds assuming the same pushing force is applied? In your solution you must show: 1. A free body diagram indicating all the forces that act on the box as well as the x and y components of all forces 2. the sum of forces for both x and y directions 3. Using the net force equations you have written down in #2 and kinematics equations as needed, show how you solve for the displacement of the box. Upload Choose a File

Answer 1

Х Question 1 R mazleg ma 0=159 masing FR TO d=0.215 imgceno mg Heres Ris Fr.is Now, F FE a Pusing quaction forice a fonce due to friction. FR =-MRL Rileumg cose - umgloso force, pa 49 Î N. x-component of the forces aching on the Fx = CP - mg sino-umgeest) Y-component of knees Fy = (R-mg eno) Here reaction fonce equal to the you 4-componet of weight bose, So, R= m Cosa i fy aog Hence net force ? = fx i t fy = (P-mg sino ung conto) to

So So here the value of acceleration is LP -masin 19-{{x9-85in 150) – (0.215414986) 15") mgsins ung can cond a= m m a= 7 a=2143 m/s2 box, u= 1 m/s taken Now the initial velo veloerty ho of the time for the displacement, fa 3 see. So, the displacement - a= ut that? @ax 2.48839) the displacement locemet of the box is (1X3) + m = 13,935 m cpl 13.935 m