Abox sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed position 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

x₂=2×1

Explanation:

Using work energy theorem, consider the gravitational potential energy at its lowest position of compression to be zero, because change in kinetic energy is 0 then;

mgx-(kx)²/2 =0   where m is mass of object, g is acceleration due to gravity, k is spring contant and x is the compression on the spring.

mgx=(kx)²/2

x=2mg/kcompression when velocity of object is Zero

But ΔK.E =-1/2mv²⇒kx²=mv² where v is velocity of the object and K.E is kinetic energy

So if kx²=mv² then

v=x *√(k/m) where v=0

Doubling v will multiply the x *√(k/m) by 2 making x₂ twice x₁

i think it’s a. this is because if they measured the mass of something with a mass of 150g, then their final answer would have to be approximately 150g. though all of their answers are almost 150, rosa and tran have the closest answers (or the exact answer) to the accurate one.

i had this question and the answer was 3.15 miles.

martina travels from school to her mom's office. from there, she walks to school. after school, she walks to the library and then to a friends house before finally ending back at home. find martina's total distance traveled.

a) 0 miles  

b) 0.4 miles  

c) 1 mile  

d) 3.15 miles