Question

For the following problem, show all work. Circle your answers. A box of mass m = 5.0 kg is sitting on a ramp, which is elevated at an angle 0 = 20°, compressing a spring Ar = 0.5 m. The spring is released, causing the box to slide up the ramp. At some point, the spring becomes detached from the box. The total length of the ramp is L = 4.0 m and it has a coefficient of kinetic friction Hk = 0.1. The spring has a spring constant k = 50 N/m. Consider the length of the spring when it is compressed to be zero; that is, assume the box's initial height off the ground is zero. a. Draw a free body diagram for the box at the moment the spring is released. Resolve forces into components as necessary. (2 pts) b. What is the initial acceleration of the block? (3 pts) c. When the block is at the top of the ramp, how much work has been done by friction? (2 pts) d. What is the box's velocity at the top of the ramp? (3 pts)

Answer 1

mo 5.0kg, 0-20', A3=0.5m Free body diagroem ,L=40m = 001,1=50N/m 0 py so from Resolve force along the incine (x-axis) and normal to the induine ly-axis. IM L 5 mg coso Alony Y-axis N = mg cosa Alony x-axi Fs - masino- tk = ma. → Kax-mgeine- Ming caso = ma. > a= kax - g (sino tulkcoso) 2=150X0.5 - 98lein 20 + 0-1X10520)] mis? [a= 6.7273 mis? When the block is at top of the ramp work done by fruction wf-Cuking Coso) Kf =-6.1X5X9.8% (0820 x4 I Wye-18.42 I using conservation of energy Wf+ klgraving & Wemangfore changes in KE. -Mengembo L - mg Leino + [Kao? = {my-o – 18:42 – 589.8 x 4X Lin20' + $x50x 80:5)= 2 xux vi? - tº-u2 – 64: 64 + / 25 = 5x 4x u . - 14 =-39.6 a since value is negative so box will not reach at the top of the ramp. Belo Before reaching at the top it will stop.