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.
Slides (d + .23) total
m g (d+.23) sin 30 = .5 k (.23)^2
Ah, you subtract .23 from .747. Thanks!
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
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
To solve this problem, we can break it down into two parts: the motion on the incline before the block hits the spring, and the motion after the block hits the spring.
Let's start by analyzing the motion on the incline before the block hits the spring.
1. Calculate the gravitational potential energy of the block:
Gravitational potential energy = mgh, where m is the mass (3.20 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the vertical height (d * sin(30°)).
Gravitational potential energy = 3.20 kg * 9.8 m/s^2 * d * sin(30°)
2. Calculate the elastic potential energy of the compressed spring:
Elastic potential energy = 0.5 * k * x^2, where k is the spring constant (443 N/m) and x is the compression of the spring (23.0 cm = 0.23 m).
Elastic potential energy = 0.5 * 443 N/m * (0.23 m)^2
Since there is no energy loss due to friction or other sources, the gravitational potential energy will be transformed entirely into elastic potential energy.
3. Equating the gravitational potential energy to the elastic potential energy, we have:
mgh = 0.5kx^2
4. Substitute the values into the equation and solve for d:
3.20 kg * 9.8 m/s^2 * d * sin(30°) = 0.5 * 443 N/m * (0.23 m)^2
Now, let's solve for d:
d = (0.5 * 443 N/m * (0.23 m)^2) / (3.20 kg * 9.8 m/s^2 * sin(30°))
After plugging in the values and using the order of operations, you should get the correct value for d.
Moving on to part (b), we need to find the distance between the point of first contact and the point where the block's speed is greatest.
The point of first contact is at a distance d from the top of the incline. The block slides an additional 23.0 cm (0.23 m) after hitting the spring before it comes to rest again momentarily. Therefore, the distance between the point of first contact and the point where the block's speed is greatest is d + 0.23 meters.
So, the distance is d + 0.23 meters.
By calculating d using the correct formula and substituting the value into the expression for the distance in part (b), you should be able to get the correct answers for both parts of the problem.