A 1.3 kg block is held against a spring of force constant 2.0×104 N/m , compressing it a distance of 0.15 m. How fast is the block moving after it is released and the spring pushes it away?

Answer 1

I think this can be most easily solved as a conservation of energy problem.

When the spring is compressed this creates potential energy (PE) that depends on the force constant (k) and the distance the spring is compressed (x). The potential energy equation is

PE = ½kx²

Plugging in k = 2 × 10⁴ N/m and x = 0.15 m we get

PE = ½(2 × 10⁴)(0.15)² = 225 J

When the block is released 225 J of potential energy gets converted to 225 J of kinetic energy (KE).

We use the equation KE = ½mv²

then we can plug in KE = 225 J, m = 1.3 kg, and solve for v.

Since KE = ½mv² this means that

v² = (2× KE)/m and

v = √[(2× KE)/m] = √[(2× 225)/1.3] = √346 = 18.6

answer (rounded to 2 significant figures)

19 m/s

Answer 2

Hi Kathy,

Notice if we equate Newton's second law with Hookes law, ma=-kx, we can solve for acceleration but NOT velocity. Therefore, we have to take another approach.

If we use energy conservation PE=KE in this case, we have 0.5 kx^2=0.5mv^2. We can solve for v.

From algebra, v=sqrt((kx^2)/m)

Sqrt((20000×0.15^2)/1.3)= 18.6 m/s.

I hope this helps!