Work done by a spring on a block when released from 14cm?

[rockchalk1312]

A block is attached to a spring which is attached to a stationary wall. When the block is pulled out to x = +5.0 cm, we must apply a force of magnitude 370 N to hold it there. We pull the block to x = 14.0 cm and then release it. How much work does the spring do on the block when the block moves from xi=+6.0 cm to (a)x= +3.0 cm, (b)x=-3.0 cm, (c)x=-6.0 cm, and (d)x=-10.0 cm?

F_s=-kd
W_s=1/2kx_i²-1/2kx_f²

Tried to find k by doing 370=-k(5)=-74
but then plugging this into:
W=1/2kxi^2-1/2kxf^2
W=1/2k(6^2)-1/2k(3^2)
W=18k-4.5k
W=13.5k
W=(13.5)(74)=999

This was definitely not the right answer. How do you go about finding k, or do you have to do something differently since the block was originally released from 14cm? Thank you!

 

[TSny]

Make sure all your numbers are in the same system of units.

 

[cepheid]

A block is attached to a spring which is attached to a stationary wall. When the block is pulled out to x = +5.0 cm, we must apply a force of magnitude 370 N to hold it there. We pull the block to x = 14.0 cm and then release it. How much work does the spring do on the block when the block moves from xi=+6.0 cm to (a)x= +3.0 cm, (b)x=-3.0 cm, (c)x=-6.0 cm, and (d)x=-10.0 cm?

F_s=-kd
W_s=1/2kx_i²-1/2kx_f²

Tried to find k by doing 370=-k(5)=-74
but then plugging this into:
W=1/2kxi^2-1/2kxf^2
W=1/2k(6^2)-1/2k(3^2)
W=18k-4.5k
W=13.5k
W=(13.5)(74)=999

This was definitely not the right answer. How do you go about finding k, or do you have to do something differently since the block was originally released from 14cm? Thank you!

Be careful with units. The distance is 5 cm. If the problem is expecting an answer in Nm, then you must make sure everything is in SI units, which means converting distances to m. Of course, if you did the rest of it right, then your answer should be in N*cm, which means it should only have been off by some factors of ten.

 

[rockchalk1312]

Of course, if you did the rest of it right, then your answer should be in N*cm, which means it should only have been off by some factors of ten.

I was off by two factors of ten. Thank you very much!

 

[Nickie]

I would approach this problem by first clarifying some details. Is the block being pulled out to x = +5.0 cm and then released, or is it being pulled out to x = 14.0 cm and then released? This will affect the initial position of the block and the corresponding spring force.

Assuming the block is being pulled out to x = +14.0 cm and then released, we can calculate the spring constant (k) using the given information. The spring force is given by F = kx, where x is the displacement from the equilibrium position. At x = +5.0 cm, the force needed to hold the block in place is 370 N, so we can write:

370 N = k(5.0 cm)
k = 74 N/cm

Now, let's calculate the work done by the spring as the block moves from xi = +6.0 cm to different positions:

(a) x = +3.0 cm:
The block moves 3.0 cm closer to the equilibrium position, so the displacement is -3.0 cm. Using the formula W = 1/2 k xi^2 - 1/2 k xf^2, we get:

W = 1/2 (74 N/cm) (6.0 cm)^2 - 1/2 (74 N/cm) (3.0 cm)^2
W = 666 J

(b) x = -3.0 cm:
The block moves 3.0 cm past the equilibrium position, so the displacement is +3.0 cm. Using the same formula, we get:

W = 1/2 (74 N/cm) (6.0 cm)^2 - 1/2 (74 N/cm) (-3.0 cm)^2
W = 666 J

(c) x = -6.0 cm:
The block moves 6.0 cm past the equilibrium position, so the displacement is +6.0 cm. Using the same formula, we get:

W = 1/2 (74 N/cm) (6.0 cm)^2 - 1/2 (74 N/cm) (-6.0 cm)^2
W = 1488 J

(d) x = -10.0 cm:
The block moves 10.0 cm past the equilibrium position, so the displacement is +10.0 cm. Using the same