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
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
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
Answer in units of cm.
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
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
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.