Question

When the block in Figure is pulled out on a frictionless surface to = 5.0 cm from its equilibrium position ( = 0), we must apply a force of magnitude 450 N to hold it there. The block is then released. How much work does the spring do on the block as the block moves from = 5.0 cm to = −3.0 cm?

I need to know how to solve it please.​

Answer 1

a) So we have

Fx=

2

1

kx 2

⇒(3.0N)x=

2

1

(50N/m)x 2

which (other than the trivial root) gives x = (3.0/25) m = 0.12 m.

(b) The work done by the applied force is W a

=Fx=(3.0N)(0.12m)=0.36J.

(c) The above equation immediately gives W s

=−W a

=−0.36J.

(d) With k f

=K considered variable and K i

=0, The equation gives K=Fx−

2

1

kx 2

. We have the derivative of K with respect to x and set the resulting expression equal to zero, in order to find the position x

c

that corresponds to a maximum value of K:x c

=

k

F

=(3.0/50)m=0.060m

We note that x

c

is also the point where the applied and spring forces “balance.”

(e) At x

c

we find K=K

max

=0.090J.

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