Forces stuff

[Cummings]

http://cummingsiam.steven-sst.com/force2.jpg

those are the questions i am having trouble with
for question 7:
work is equal to Fscos()
where
f = 100 n
s = 1cm = .01m
cos() = cos(0) = 1

work = 100 * .01 * 1
work = 1 joul?

[KLscilevothma]

http://cummingsiam.steven-sst.com/forces2.jpg
It is a dead link.

[Cummings]

should be fixed now..i added an extra s its force2.jpc not forces2.jpg

[Dave]

7a
100N
7c
2W
7b
Bookwork.

Yes, work = one Joule.

[KLscilevothma]

The graph given is a F(force)-s(compression) graph. Work done equals area under the straight line. So it isn't 100*0.1*1

[Dave]

Originally posted by KL Kam
The graph given is a F(force)-s(compression) graph. Work done equals area under the straight line. So it isn't 100*0.1*1

The graph isn't in SI units tho.

[KLscilevothma]

we can change the scale of the x-axis without affecting the result.

Work done is still area under the straingt line.
therefore work done = (100)*(0.01)*(0.5)= 0.5 J

7a
100N
Also unit for work done is Joule, not Newton(N)

[Cummings]

Originally posted by KL Kam
we can change the scale of the x-axis without affecting the result.

Work done is still area under the straingt line.
therefore work done = (100)*(0.01)*(0.5)= 0.5 J


Also unit for work done is Joule, not Newton(N)
yes, its .5J fpr 7a and 1w for 7c
i have never worked with force-compression graphs before..but now i know.

Question 8 i can't do either...
i don't know the relationship between speed and force...
force = mass * acceleration is all i know

[KLscilevothma]

#8
Try to use "energy approach". Kinetic energy is changed to work done to compress the ball. (assume there's no energy loss)

[Cummings]

well, its kinetic energy is .5 * mass * velocity squared
so .5 * .1(100 grams) * 100(10 squared)
= 5 jouls

now..work is measured in jouls..

now..the graph is linear so we can change the scale to use X to represent both force and compression.

5j = force * compression *.5
10 = force * compression
let X = force and compression
10 = X squared
x= square root of 10 = 3.2 (rounded)
this equals 3.2cm for compression - and converting back our scale we get 320N for our force.
Thus the maximum compression when the ball hits a wall at 10ms-1 is 32mm

woooo..that was some tricky stuff.
But its right.
320 * .032 * .5 = 5j
5 = .5 * .1 * v squared
v squared = 100
v = 10ms

its all correct :)
Thanks