Answer:
A) K / K₀ = 4 b) v / v₀ = 4
Explanation:
A) For this exercise we can use the conservation of mechanical energy
in the problem it indicates that the displacement was doubled (x = 2xo)
starting point. At the position of maximum displacement
Em₀ = Ke = ½ k (2x₀)²
final point. In the equilibrium position
= K = ½ m v²
Em₀ = Em_{f}
½ k 4 x₀² = K
(½ K x₀²) = K₀
K = 4 K₀
K / K₀ = 4
B) the speed value
½ k 4 x₀² = ½ m v²
v = 4 (k / m) x₀
if we call
v₀ = k / m x₀
v = 4 v₀
v / v₀ = 4
Answer:
120 J
Explanation:
KE = mv²/2 = (0.15 kg * [40 m/s]²)/2 = 120 J
Given:
1st run: 20 meters North
2nd run: 15 meters East
time: 15 seconds
Average speed = total distance covered / total time taken
Ave. Speed = (20m + 15m) / 15s
Ave. Speed = 35m / 15s
Ave. Speed = 2 1/3 meters per second
1) newton of the surfaces in contact (the smoothness of the surfaces)
2) how hard the surfaces press together
Work Done = Force x Distance Moved
Work Done = 25 x 15 = 375 Joules