Please show work and explain answers to be understandable A 14.0 kg stone slides
ID: 1360456 • Letter: P
Question
Please show work and explain answers to be understandable
A 14.0 kg stone slides down a snow-covered hill (the figure(Figure 1), leaving point A with a speed of 12.0 m/s . There is no friction on the hill between points A and B. but there is friction on the level ground at the bottom of the hill, between B and the wall. After entering the rough horizontal region, the stone travels 100 m and then runs into a very long, light spring with force constant 2.00 N/m . The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80. respectively. What is the speed of the stone when it reaches point B? How far will the stone compress the spring? Will the stone move again after it has been stopped by the spring?Explanation / Answer
a.
The speed when it reaches the B depends on the vertical height alone
v^2 = 12^2 +19.6*20 = 144 + 392 (using the equation v^2 = u^2 + 2as)
v = 23.15 m/s
b. The acceleration due to friction is -ug.
The velocity after travelling 100 m is given by
v^2 = u^2 - 2as = (144+ 392) – (2*0.2*9.8*100) = 144. (Using kinetic friction since it is in motion)
If the spring is compressed to a distance s, (Using energy principle)
v^2 = 2ugs + (k/m) s^2
144 = (2*0.2*9.8*s) + (2/14) s^2
s = 20.86 m
c) The static frictional acceleration is = - 0.8*9.8 =-7.84
The acceleration due to the spring force is = 2*20.86/14 = 2.98 m/s^2
Since the maximum frictional acceleration is greater than 2.98 m/s^2, the mass will not move further.