- #1
pdonovan
- 17
- 0
Question:
You and your friend Peter are putting new shingles on a roof pitched at 25 . You're sitting on the very top of the roof when Peter, who is at the edge of the roof directly below you, 5.5 away, asks you for the box of nails. Rather than carry the 2.0 box of nails down to Peter, you decide to give the box a push and have it slide down to him. If the coefficient of kinetic friction between the box and the roof is 0.55, with what speed should you push the box to have it gently come to rest right at the edge of the roof?
My attempt:
I drew a free body diagram for the box while it is in motion. From there I concluded that:
Fnet = Fgsin20 - f
--> mgsin20 - .55N = ma
--> 2(9.8)sin20 - .55(2*9.8*cos20) = 2a
--> a = -1.7131
I then used the kinematics equations Vf = Vi + aT; and Sf = ViT + .5a(T^2) to find Vi = 4.34m/s.
Any help or guidance is greatly appreciated! Thank you.
You and your friend Peter are putting new shingles on a roof pitched at 25 . You're sitting on the very top of the roof when Peter, who is at the edge of the roof directly below you, 5.5 away, asks you for the box of nails. Rather than carry the 2.0 box of nails down to Peter, you decide to give the box a push and have it slide down to him. If the coefficient of kinetic friction between the box and the roof is 0.55, with what speed should you push the box to have it gently come to rest right at the edge of the roof?
My attempt:
I drew a free body diagram for the box while it is in motion. From there I concluded that:
Fnet = Fgsin20 - f
--> mgsin20 - .55N = ma
--> 2(9.8)sin20 - .55(2*9.8*cos20) = 2a
--> a = -1.7131
I then used the kinematics equations Vf = Vi + aT; and Sf = ViT + .5a(T^2) to find Vi = 4.34m/s.
Any help or guidance is greatly appreciated! Thank you.