how far up the incline does it travel before coming to rest and then sliding back down?

Question

A spring with k = 10.0 N/m is at the base of a frictionless 30.0° inclined plane. A 0.50 kg object is pressed against the spring, compressing it 0.3 m from its equilibrium position. The object is then released. If the object is not attached to the spring, how far up the incline does it travel before coming to rest and then sliding back down? (See the figure below.)
_________m

Can someone show me how to solve this in steps? I've tried many times and I can't seem to get the right answer. Thanks!

Answer 1

Use conservation of energy. The energy stored in the spring when compressed is 0.5*k*∆x^2. ∆x is the amount of compression (0.3 m). This energy will be converted to potential energy of the mass at its highest elevation above the starting point. That potential energy is m*g*h; Equating these gives the relation

m*g*h = 0.5*k*∆x^2

solve for h.

The distance along the plane is h / sinø, where ø is the angle of inclination.

Answer 2

We don't need to work with force, just acceleration, as force and acceleration are proportional with respect to mass and our mass is constant. Anyway, the frictional acceleration acting on the block on the incline would be... Friction=μgcos(θ) (We're taking the x component of gravity on the incline, thus cosine). The gravitational acceleration on the incline (which acts in the same direction) would be... Gravity=gsin(θ) (We're taking the y component of gravity on the incline, thus sine). So the sum of the acceleration acting on the block, Σa = μgcos(θ) + gsin(θ), this is equal to a constant value, we'll call it S. S = μgcos(θ) + gsin(θ) = (0.180) * (9.8 m/s^2) * cos(14 degrees) + (9.8 m/s^2) * sin(14 degrees) = 4.08243624 m/s^2 Now that we have acceleration, we can work using conservation of energy. As mass is irrelevant and always the same + not given, we can use a less known unit called Grays (Gy, J/kg or m^2/s^2). Now kinetic energy is equal to K = .5mv^2. Kinetic Grays would therefore be equivalent to KGy = .5v^2. This is our initial Grays. The loss of Kinetic Grays on the incline by the block is equal to LKGy = a*d = S*d...where d is the distance travelled on the incline (This comes from W=f*d), thus we can set the two equal. S*d = .5*v^2 d = .5*v^2/S d = .5*((1.00 m/s)^2)/(4.08243624 m/s^2) d = 12.2475887 centimeters So the answer would be .12 m, maybe you should check if you got units right?

Answer 3

The secret is knowing how much energy is stored in the spring =kxx/2=10*.3*.3/2
=.45J Since the plane is frictionless, all this energy is = to mgh where h is the height the object reaches. .45J=.5*9.8h
h=.092m. h=dsin30 where d is the distance up the ramp. So
d=h/sin30=.184m