Mass-spring system on an incline

[kris8969]

A spring is mounted at an angle of theta = 39degrees on a frictionless incline as illustrated in the figure below. The spring is compressed to 15 cm where it is allowed to propel a mass of 4.9 kg up the incline.

(a) If the spring constant is 580 N/m, how fast is the mass moving when leaves the spring?
m/s
[5 points] 5 attempt(s) made (maximum allowed for credit = 5)
[after that, multiply credit by 0.5 up to 10 attempts]
1.864 NO

(b) To what maximum distance from the starting point will the mass rise up the incline?
m

Homework Equations

i have no clue what is correct because i have tried so many different ones

The Attempt at a Solution

but for some reason this is not the correct answer

Step One
=======
Find the Force created by the spring
F = kx
k = 580 N/m
x = 15cm = 15*[1 m/100 cm] = 0.15 m

F = 580*0.15 = 87 N

Step Two
=======
Find the force created (in opposition) by the mass trying to slide down the incline.

The formula for the force trying to go down the incline is F = mg*sin(A)
A = 39o
m = 4.9 kg
g = 9.81 m/s^2

F-incline = 4.9*9.81 * sin(28)
F-incline = 30.25 N

Step Three
========
Find the net upward force created by the spring.
F-net = F-spring - F-incline
F-net = 87 - 30.25= 56.75

Step Four
=======
Find the acceleration.
F = m*a
56.75 = 4.9*a
11.58 m/s^2 = a

Step Four
=======
Find the final velocity of the mass as it departs from the spring.
vi = 0
a = 11.58
d = 0.15 m
vf = ??

vf^2 = vi^2 + 2*a*d
vf^2 = 0 + 2*11.58*0.15
vf^2 = 3.47
vf = 1.86 m/s

 

[PhanthomJay]

It is easier to use work energy methods, but your method is OK, except for the value you are using for the spring force. The spring force is 87 N when it is fully compressed, but then decreases linearly to zero when it returns to its unstretched length, so you have to use the average spring force in your equations. Is the incline at 39 degrees or 28 degrees??

 

[kris8969]

the angle is 39^o

but are you saying for the first equation where i use F=kx, i should use (580+0)/2 for the value of k instead of just 580?

 

[PhanthomJay]

the angle is 39^o

but are you saying for the first equation where i use F=kx, i should use (580+0)/2 for the value of k instead of just 580?

No, the value of k is what it is...580 N/m. The force of the spring on the mass is F_s = kx, where x varies from x=0.15 m when it is compressed at the starting position (F_s = 87 N), to x = 0 when the mass leaves the spring (F_s =0 at that point). So the average force exerted by the spring on the mass is ? Use that value as the spring force when applying Newton's 2nd law.

 

[rwisz]

(a) The mass will be moving at a speed of 1.86 m/s as it leaves the spring.

(b) To find the maximum distance the mass will rise, we can use the formula for projectile motion:
d = vi^2*sin(2A)/g

vi = 1.86 m/s
A = 39o
g = 9.81 m/s^2

d = 1.86^2*sin(78)/9.81
d = 0.74 m

Therefore, the maximum distance the mass will rise up the incline is 0.74 meters.