Question

The spring in the figure below has a spring constant of 1150 N/. It is compressed 28 cm, then launches a 200 g block. The horizontal surface is frictionless, but the block's coefficient of kinetic friction on the incline is 0.20. What distance d does the block sail through the air? Enter a numbor 2.0 m 200g 5°

Answer 1

2oogm Spring Constant - 1150Nlm =k, haam Comprension in spring = 28cm = mass of block - 200 grom Froosh -) 0.2kg=m Cofficient of kinetic friction, d. on incline= 0.2 So, I sinus-2 ud la 2.828 m. fo, Total Stored energy in Spring = increase in potential energy & Energy lost in frictions et kinetic energy ato. So, I ke² = mgh - lamgl + 1 m² tx1150 x(0.28) = 0.2x9.842 -02 X0-2X9.8 x t 1 x 0.2 x 2.828 2. us.08 = 3.92 – 1.10857 toolv? 42.268 = 0.122 ) 2 = 422.685 . V = 20.559 - 20.56 m/s Now, It becomes projectile with velocity velocity = 20.56 mis making an angle 45° with axis.

20.56m's Dyro So, Here d= Range of this stro projectile. So, d = usinzo - 422.685 x 12 9.8 1430 131 m A 43.131m te fin 20-singon = 1