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docker41
1 month ago
8

A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit

ion while momentarily coming to rest.
1)If the initial speed of the box were doubled, how far x2 would the spring compress?

x2 = ?2 x1
x2 = 2 x1
x2 = 4 x1
Physics
1 answer:
serg [3.5K]1 month ago
7 0

Answer:

x₂=2×1

Explanation:

According to the work-energy theorem, we can assume that the gravitational potential energy at the lowest point of compression is zero since the kinetic energy change is 0;

mgx-(kx)²/2 =0 where m refers to the object's mass, g indicates the acceleration due to gravity, k denotes spring constant, and x represents the spring's compression.

mgx=(kx)²/2

x=2mg/k----------------compression when the object is at rest

However, ΔK.E =-1/2mv²⇒kx²=mv² -----------where v symbolizes the object's velocity and K.E signifies kinetic energy

Thus, if kx²=mv² then

v=x *√(k/m) ----------------where v=0

<pDoubling v results in multiplying x *√(k/m) by 2, leading to x₂ being double x₁

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