Answer all parts to the question please. I have attached a copy of the only formulas we use to solve problems
![Position, Velocity, Acceleration<br>Rectangular Coordinate:<br>i = xỉ + yj + zk<br>ü = v,i + v,j + v,k = xi + ÿj + żk<br>a = a,i + a,j + a,k = xï + ÿj+ žk<br>Projectile Motion<br>a = -gj<br>v = vĩ + v,j = voxī+(-gt + voy)j<br>Kinematic Equations for Two Points on the Same<br>Rigid Body:<br>%3D<br>/A<br>Pure Rolling:<br>vc = rw; ac = ra, consistent with the rolling direction;<br>v¡ = 0; a¡ = rw² toward C<br>7 = xĩ + yỹ = (Voxt + xo)ï + (–gt2 + voyt + Yo)j Eq. of Motion for a Rigid Body in Planar Motion:<br>1<br>gt²<br>ΣF -<br>Tangent and Normal Coordinate:<br>* = 7(s)<br>i = vũ; = sū;<br>à = a,ū, + a„īn = šū, +ūn<br>= māg<br>ΣΜ 1ς α<br>ΣΜΟ-10α<br>EMQ = Iga + (Fc/ × mãc),<br>= lọa + (Tc/Q × māo), Q: any point<br>G: mass center<br>O: pivot point or instantaneous center<br>213/2<br>1+]<br>|d²y|<br>p =<br>Kinetic Energy of a Rigid Body<br>dx2<br>T =mv,² +Igw²<br>2<br>Cylindrical Coordinate:<br>= rū, + zū,<br>v = v,ũ, + vgũg + v,ū, = rū, + rôūo + żū,<br>à = a,ūr + agūg + azūz<br>= (* – rô?)ũ, + (rö + 2řė)ūg + žū,<br>T =<br>Work Done on a Rigid Body<br>- [ Fds + f Mdo<br>U =<br>Newton's Second Law:<br>2F = mã<br>Work-Energy Principle:<br>T; +U.-2 = T2<br>1<br>kinetic energy: T =<br>mv²<br>work done: U =<br>F• dr<br>*work of the force exerted by a spring:<br>U1-2 = k(sỉ – s3)<br>*work of the force exerted by friction:<br>U1-2 = -fd<br>Conservative Force Field:<br>T1 + Vị = T2 + V2<br>T: kinetic energy<br>V:potential energy<br>](https://s3.us-east-1.amazonaws.com/storage.unifolks.com/qimg-008/008_wnflzp0-35dcy33i.jpeg)
Extracted text: Position, Velocity, Acceleration Rectangular Coordinate: i = xỉ + yj + zk ü = v,i + v,j + v,k = xi + ÿj + żk a = a,i + a,j + a,k = xï + ÿj+ žk Projectile Motion a = -gj v = vĩ + v,j = voxī+(-gt + voy)j Kinematic Equations for Two Points on the Same Rigid Body: %3D /A Pure Rolling: vc = rw; ac = ra, consistent with the rolling direction; v¡ = 0; a¡ = rw² toward C 7 = xĩ + yỹ = (Voxt + xo)ï + (–gt2 + voyt + Yo)j Eq. of Motion for a Rigid Body in Planar Motion: 1 gt² ΣF - Tangent and Normal Coordinate: * = 7(s) i = vũ; = sū; à = a,ū, + a„īn = šū, +ūn = māg ΣΜ 1ς α ΣΜΟ-10α EMQ = Iga + (Fc/ × mãc), = lọa + (Tc/Q × māo), Q: any point G: mass center O: pivot point or instantaneous center 213/2 1+] |d²y| p = Kinetic Energy of a Rigid Body dx2 T =mv,² +Igw² 2 Cylindrical Coordinate: = rū, + zū, v = v,ũ, + vgũg + v,ū, = rū, + rôūo + żū, à = a,ūr + agūg + azūz = (* – rô?)ũ, + (rö + 2řė)ūg + žū, T = Work Done on a Rigid Body - [ Fds + f Mdo U = Newton's Second Law: 2F = mã Work-Energy Principle: T; +U.-2 = T2 1 kinetic energy: T = mv² work done: U = F• dr *work of the force exerted by a spring: U1-2 = k(sỉ – s3) *work of the force exerted by friction: U1-2 = -fd Conservative Force Field: T1 + Vị = T2 + V2 T: kinetic energy V:potential energy
![The package has a mass of 0.3 kg and slides down the chute from point A at rest. If the kinetic friction<br>coefficient ug = 0.2 on the straight section AB, (a) determine the velocity of the package when it reaches point<br>B [You may use the Newton's 2nd Law or the energy method]. If the curved section BC is frictionless, use the<br>energy method to determine (b) the velocity of the package at point C. Then, determine (c) the reaction force<br>exerted on the package from the track at point C. Also, detemine (d) the horizontal distance d where the<br>package falls on the ground at D.<br>45<br>02-<br>45°<br>6.2 m<br>в<br>32 m<br>[8] (a) v»<br>(7] (b) vc<br>[5] (c) Fy =<br>[5](d) d =<br>5.6 m<br>](https://s3.us-east-1.amazonaws.com/storage.unifolks.com/qimg-008/008_e9jlzp0-pdzokk2c.jpeg)
Extracted text: The package has a mass of 0.3 kg and slides down the chute from point A at rest. If the kinetic friction coefficient ug = 0.2 on the straight section AB, (a) determine the velocity of the package when it reaches point B [You may use the Newton's 2nd Law or the energy method]. If the curved section BC is frictionless, use the energy method to determine (b) the velocity of the package at point C. Then, determine (c) the reaction force exerted on the package from the track at point C. Also, detemine (d) the horizontal distance d where the package falls on the ground at D. 45 02- 45° 6.2 m в 32 m [8] (a) v» (7] (b) vc [5] (c) Fy = [5](d) d = 5.6 m