Solving Spring/Work Problem: Find Block's Movement on Incline

[Fert]

This is the problem:
A 235g block is pressed against a spring (with K=1.32KN/M) until the block compresses the spring 10.0cm. The spring rests at the bottom of a ramp inclined at 53.4deg to the horizontal.

A) Using energy considerations, determine how far up the incline the block moves before it stops if there is no friction between the block and the ramp.


I've draw a free body diagram and setup the problem like this,

Forces in x dir: Force of spring(Fs)-MgSIN(53.4)
Forces in y dir: Normal (N)- MgCOS(53.4)=0

Work due to spring-work due to gravity = K(final)-K(inital) , K(inital)=0 since V=0

So, (1/2)Kx^2 -(MgSIN(53.4))(x) =1/2 mV^2 (where x = .10 meters)

I solve for V to find out the speed at the equilibrium point and then use:

V^2(final)= V^2(inital) + 2ad to find the distance it travels after that,
(I use a= -gSin(53.4) for this,)

anyways I keep getting the wrong answer and I don't know where I went wrong? I'd really appericiate if someone could show me where I'm getting messed up.

P.S. sorry for the long post I'm not really sure how to write a lot of this stuff out on the computer.

 

[member 553083]

The answer you are looking for is 0.18 m. The mistake you are making is that you are trying to solve for the velocity at the equilibrium point, which is not necessary. All you need to do is use conservation of energy. The potential energy at the start is the work done by the spring plus the work done by gravity:PE = (1/2)Kx^2 + MgSIN(53.4)x The kinetic energy at the end is half the mass times the velocity squared:KE = (1/2)mv^2Since energy is conserved, PE = KE, so you can equate them and solve for v, then plug that into the equation for motion with constant acceleration: v^2 = 2as where a = -gSIN(53.4). This will give you the distance traveled.

 

[wais]

To solve this problem, we need to use the conservation of energy principle, which states that the total energy of a system remains constant. In this case, the initial energy of the system is only in the form of potential energy stored in the compressed spring, and the final energy is in the form of both potential and kinetic energy of the block.

First, let's calculate the potential energy stored in the compressed spring:

U = (1/2)kx^2 = (1/2)(1.32 kN/m)(0.1 m)^2 = 0.0066 kN·m

Next, we need to find the final kinetic energy of the block, which is given by:

Kf = (1/2)mv^2

To find the final velocity, we can use the conservation of energy principle again:

U = Kf

0.0066 kN·m = (1/2)mv^2

v = √(2U/m) = √(2(0.0066 kN·m)/(0.235 kg)) = 0.220 m/s

Now, we can use the equation V^2(final) = V^2(initial) + 2ad to find the distance the block travels after it leaves the spring. Since the block is moving up the incline, the acceleration will be in the opposite direction of the displacement (up the incline), so we can use a = -gSin(53.4):

0 = (0.220 m/s)^2 + 2(-9.8 m/s^2)Δd

Δd = 0.024 m

Therefore, the block will travel 0.024 m up the incline before coming to a stop.