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) Let h be the height the block of mass attains so that h = d sin

Now, the potential energy stored in the spring when it is compressed, is gained by the block in the form of potential energy as it reaches height h.

(1/2) k x2 = m g h

(0.5) k x2 = m g d sin

d = [ (0.5) k x2 ] / [ m g sin ]

d = [ (0.5) (150 N/m) (0.10 m)2 ] / [ (0.4 kg) (9.8 m/s2) sin(30o) ]

d = 3.83 m

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