Mechancics MEI question Stuck

A mechanism for firing a ball in a table-top game is shown in the diagrams. A moving piston of mass 0.1 kg slides freely in a barrel, which is fixed. A spring of natural length 0.15 m and negligible mass is fitted inside the barrel so that before being primed for use the configuration is that shown in the diagram below. You may assume that the force due to the compression of the spring may be modelled by Hooke's Law with a stiffness of 200 Nm-1.

Show that the energy stored in the spring is 0.25 J when it is in the configuration above.

This is the first question i was a able to get this segment

The mechanism is primed for firing by pulling the piston back a distance d,in metres, where d <0.1, as shown in the diagram below.

Show that the spring has been compressed by a distance (d+ 0.05) m and that the work done in priming the mechanism is 10d(l + 10d).

The first part to this question is relatively straightforward but i cant find the solution to work done

Reply 1

LeoNGUYEN:)

OP

5

Question 4ii

Hooke's Law assessment.pdf147.7KB

Reply 2

mqb2766

21

What is the work done formula?

Reply 3

LeoNGUYEN:)

OP

5

work done is the integral for force as the force is changing so the formula is

(Kx^2)/2
where k is spring constant and x is extention

what i did:

(200(d+0.5)^2)/2 - (200(0.5)^2)/2
which i then got
100(d^2+d) which is not the requested 10d(1+10d)

(edited 3 years ago)

Reply 4

mqb2766

21

Original post by LeoNGUYEN:)

work done is the integral for force as the force is changing so the formula is

(Kx^2)/2
where k is spring constant and x is extention

So what do you get (show working)?

Reply 5

LeoNGUYEN:)

OP

5

Original post by LeoNGUYEN:)

work done is the integral for force as the force is changing so the formula is

(Kx^2)/2
where k is spring constant and x is extention

what i did:

(200(d+0.5)^2)/2 - (200(0.5)^2)/2
which i then got
100(d^2+d) which is not the requested 10d(1+10d)

^^^ this

Reply 6

mqb2766

21

Original post by LeoNGUYEN:)

work done is the integral for force as the force is changing so the formula is

(Kx^2)/2
where k is spring constant and x is extention

what i did:

(200(d+0.5)^2)/2 - (200(0.5)^2)/2
which i then got
100(d^2+d) which is not the requested 10d(1+10d)

Is it 0.05

Reply 7

DFranklin

18

Original post by LeoNGUYEN:)

work done is the integral for force as the force is changing so the formula is

(Kx^2)/2
where k is spring constant and x is extention

what i did:

(200(d+0.5)^2)/2 - (200(0.5)^2)/2
which i then got
100(d^2+d) which is not the requested 10d(1+10d)

Shouldn't it be d+0.05 not d+0.5?