## Work and Kinetic Energy Problem involving friction hw problem, DUE SOON

1. The problem statement, all variables and given/known data

Part 1.

A(n) 2300 g block is pushed by an external
force against a spring (with a 22 N/cm spring
constant) until the spring is compressed by
15 cm from its uncompressed length. The
compressed spring and block rests at the bottom of an incline of 32

.
The acceleration of gravity is 9.8 m/s
2
.
Note: The spring lies along the surface of
the ramp (see ﬁgure).
Assume: The ramp is frictionless.
Now, the external force is rapidly removed
so that the compressed spring can push up
the mass.

How far up the ramp (i.e., ℓ the length
along the ramp) will the block move (measured from the end of the spring when the
spring is uncompressed) before reversing direction and sliding back?
Remember, the block is not attached to the
spring.
Answer in units of cm.

Part 2.

Repeat the ﬁrst part of this question, but
now assume that the ramp has a coeﬃcient of
friction of  = 0.423 .
Keep all other assumptions the same.
Answer in units of cm

2. Relevant equations

Without friction = Ui + Ki = Uf + Kf, where in this problem Ui = 1/2kx^2, Ki = 0, Uf = mgh and Kf = 0. So 1/2 kx^2 = mgh

With friction = Ui + Ki + (-Fk*l) = Uf + Kf

So 1/2kx^2 - mu*(mgcosθ)*l = mgh

3. The attempt at a solution

I figured out Part 1.

From the equation above

h = kx^2/2mg

l = h/sin32

Then I subtracted .15 m and converted to cm. I got the correct answer. When I tried to solve for the length with the added friction in, I first forgot to account for the length in (mu*(mgcosθ*l).

I remembered to add in l, but I think I'm using the wrong l. The first l I used was .45 because that is the length of the ramp. I then tried .30 ( because .45-.15)

Here is my current equation and answer but I am not sure if I chose the correct l and therefore I don't know if it is correct. Please advise.

2200 N/m *.15^2m - .423(2.3 kg * 9.8m/s^2*cos32°*.30/2*2.3kg*9.8 m/2^2

h = 1.044239263 m

l = h/sinθ

so: 1.044239263 m/sin32 = 1.97056294 m

1.97056294 m - .15 m = 1.821 m

1.821 m * 100 = 182.1 cm

Sorry for not having a picture.
 PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity

Recognitions:
Homework Help
 Quote by emh182 1. The problem statement, all variables and given/known data Part 1. A(n) 2300 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed by 15 cm from its uncompressed length. The compressed spring and block rests at the bottom of an incline of 32 ◦ . The acceleration of gravity is 9.8 m/s 2 . Note: The spring lies along the surface of the ramp (see ﬁgure). Assume: The ramp is frictionless. Now, the external force is rapidly removed so that the compressed spring can push up the mass. How far up the ramp (i.e., ℓ the length along the ramp) will the block move (measured from the end of the spring when the spring is uncompressed) before reversing direction and sliding back? Remember, the block is not attached to the spring. Answer in units of cm. Part 2. Repeat the ﬁrst part of this question, but now assume that the ramp has a coeﬃcient of friction of  = 0.423 . Keep all other assumptions the same. Answer in units of cm 2. Relevant equations Without friction = Ui + Ki = Uf + Kf, where in this problem Ui = 1/2kx^2, Ki = 0, Uf = mgh and Kf = 0. So 1/2 kx^2 = mgh With friction = Ui + Ki + (-Fk*l) = Uf + Kf So 1/2kx^2 - mu*(mgcosθ)*l = mgh 3. The attempt at a solution I figured out Part 1. From the equation above h = kx^2/2mg l = h/sin32 Then I subtracted .15 m and converted to cm. I got the correct answer. When I tried to solve for the length with the added friction in, I first forgot to account for the length in (mu*(mgcosθ*l). I remembered to add in l, but I think I'm using the wrong l. The first l I used was .45 because that is the length of the ramp. I then tried .30 ( because .45-.15) Here is my current equation and answer but I am not sure if I chose the correct l and therefore I don't know if it is correct. Please advise. 2200 N/m *.15^2m - .423(2.3 kg * 9.8m/s^2*cos32°*.30/2*2.3kg*9.8 m/2^2 h = 1.044239263 m l = h/sinθ so: 1.044239263 m/sin32 = 1.97056294 m 1.97056294 m - .15 m = 1.821 m 1.821 m * 100 = 182.1 cm Sorry for not having a picture.
The two bits I highlighted in red look like you have left out the half in your calculation. Perhaps you had multiplied through the calculation by two before substituting???

I must say I find this "technique" of including the units with every number while you substitute most distracting [though I have seen other people doing the same], making it all but impossible to follow the calculation to see if you are doing the right thing. It is also a habit that is foreign to anything I have seen in a school or university here where I live.

 Similar discussions for: Work and Kinetic Energy Problem involving friction hw problem, DUE SOON Thread Forum Replies Introductory Physics Homework 1 Introductory Physics Homework 3 Introductory Physics Homework 3 Introductory Physics Homework 4 Introductory Physics Homework 10