Energy work calculations (please check over)

[blackout85]

A 2-kg block is attached to a horizontal ideal spring with a spring constant of 200N/m. When the spring has its equilibrium length the block is given a speed of 5 m/s. What is the maximum elongation of the spring?

1/2mv= 1/2KX^2
max PE equals starting KE.
then follows
x= v/sqrt k
x=.25 , that is the answer I am getting. But how is the book getting .05m

my second question:
A 6.0kg block is released from rest 80m above the ground. When it has fallen 60m its kinetic energy is:
I got 1200J as my answer
mgh=(6.0*9.81* 60m)=3531.6
PE + KE= (ME + KE)
3531.6 + KE= (6.0*9.81*80m)
KE= 1177.2
is that right

 

[PhanthomJay]

A 2-kg block is attached to a horizontal ideal spring with a spring constant of 200N/m. When the spring has its equilibrium length the block is given a speed of 5 m/s. What is the maximum elongation of the spring?

1/2mv= 1/2KX^2
max PE equals starting KE.
then follows
x= v/sqrt k
x=.25 , that is the answer I am getting. But how is the book getting .05m

my second question:
A 6.0kg block is released from rest 80m above the ground. When it has fallen 60m its kinetic energy is:
I got 1200J as my answer
mgh=(6.0*9.81* 60m)=3531.6
PE + KE= (ME + KE)
3531.6 + KE= (6.0*9.81*80m)
KE= 1177.2
is that right

You posted #1 earlier. your method is correct, but your equation and/or math is incorrect.
1/2mv^2 = 1/2kx^2
mv^2 = kx^2
mv^2/k = x^2
2(5)(5)/200 = x^2 = 0.25
x = 0.5 m (I don't know why the book says .05m).

for #2 its
PE_initial + KE_initial = PE_final + KE_final
For simplicity, choose the 60m point as the reference height for PE (PE =0 at this point). There will then be a couple of 0 terms in that equation, and you can easily solve for v, without worrying about the 80 m height at all..

 

[kyle8921]

I would first double check the calculations to ensure accuracy. For the first question, the equation used (1/2mv= 1/2KX^2) is the correct formula for calculating the maximum elongation of a spring. The answer of 0.25m is also correct, as it matches the units of the given speed (m/s). It is possible that the book made a mistake in their calculations or used a different formula.

For the second question, the answer of 1200J is correct. The equation used (mgh=PE+KE) is also correct. It is important to note that the potential energy (PE) at the starting point (80m) should be subtracted from the total potential energy at 60m, as the block has already fallen 60m. This would result in a potential energy of 1764.48J at 60m. When added to the kinetic energy of 1177.2J, the total mechanical energy at 60m is 2941.68J, which is the same as the total mechanical energy at the starting point (80m). Therefore, the kinetic energy at 60m is 1200J. It is possible that the book may have made a mistake in their calculations or used a different formula.