An air-track glider (m = 0.4 kg) is pushed against a Hooke\'s-Law spring (k= 150

Question

An air-track glider (m = 0.4 kg) is pushed against a Hooke's-Law spring (k= 150 N/m), compressing the spring by 10 cm. The glider is then released and the glider coasts upward along the slanted track. How far along the track does the glider travel before starting back down? The potential energy of a 4-kg particle is given by U(x, y, z) = -(5 N/m)x^2 - (4 N/m^2)y^3. (Assume there is no gravitational force, and this is its total potential energy.) What are the x-,y-, and z-components of the force acting on this particle? If the particle is released from rest at the point (x_0, y_0, z_0)= (0, +2 m, 0), what is its speed after traveling a distance of 5 m?

Explanation / Answer

1)
k = 150 N/m
m = 0.4 Kg
x = 10 cm = 0.1m

Using Energy Conservation,
Initial Spring Potential Energy = Final Gravitational Potential Energy
1/2*kx^2 = m*g*h
1/2*150*0.1^2 = 0.4*9.8*h
h = 0.191 m

Now,
Sin(30) = h/d
Sin(30) = 0.191/d
d = 0.382 m = 38.2 cm

But the Block was initially Compressed by 10 cm
Therefore total Distance moved, d = 28.2 cm

please post seperate questions in sepearate post only.

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