Slides (d + .23) total
m g (d+.23) sin 30 = .5 k (.23)^2
A 3.20 kg block starts at rest and slides a distance d down a frictionless 30.0° incline, where it runs into a spring (Fig. 8-6). The block slides an additional 23.0 cm before it is brought to rest momentarily by compressing the spring, whose spring constant k is 443 N/m.
Fig. 8-6
(a) What is the value of d?
(b) What is the distance between the point of first contact and the point where the block's speed is greatest?
I have tried this:
mgh = .5kx^2
mg(d*sin30) = .5kx^2
d = kx^2/2mgsin30
d = 443(.23)^2 / 2(3.2)(9.8)sin30
d = .747, but it is not correct
How do I solve this problem?
Thanks.
6 answers
Ah, you subtract .23 from .747. Thanks!
But what about part b? Do you find when net work = 0?
But what about part b? Do you find when net work = 0?
Not that easy
m g (d+.23) sin 30 = .5 k (.23)^2
sin 30 = .5 so
m g (d+.23)= k (.23)^2
3.2 (9.8)(d+.23) = 443 * .23^2
===========================
for part b
m g (d+x) = k (x)^2 + .5 m v^2
we know d from part a
.5 m v^2 = m g d + m g x - k x^2
m v dv/dx = m g - 2 k x
dv/dx = 0 at max v
2 k x = m g
LOL - balances the weight component
m g (d+.23) sin 30 = .5 k (.23)^2
sin 30 = .5 so
m g (d+.23)= k (.23)^2
3.2 (9.8)(d+.23) = 443 * .23^2
===========================
for part b
m g (d+x) = k (x)^2 + .5 m v^2
we know d from part a
.5 m v^2 = m g d + m g x - k x^2
m v dv/dx = m g - 2 k x
dv/dx = 0 at max v
2 k x = m g
LOL - balances the weight component
Thank you for your help!
You are welcome.
for part b
.5 m g (d+x) = .5 k (x)^2 + .5 m v^2
we know d from part a
m v^2 = m g d + m g x - k x^2
m v dv/dx = m g - k x
dv/dx = 0 at max v
k x = m g
.5 m g (d+x) = .5 k (x)^2 + .5 m v^2
we know d from part a
m v^2 = m g d + m g x - k x^2
m v dv/dx = m g - k x
dv/dx = 0 at max v
k x = m g
2 answers