An inclined plane of angle theta = 20. 0 degree has a spring of force constant k

Question

An inclined plane of angle theta = 20. 0 degree has a spring of force constant k = 460 N/m fastened securely at the bottom so that the spring is parallel to the surface as shown the figure below. A block of mass m = 2. 33 kg is placed on the plane at a distance d = 0. 282 m from the spring. From this position, the block is projected downward toward the spring with speed v = 0. 750 m/s. By what distance is the spring compressed when the block momentarily comes to rest? differs from the correct answer by more than 100%. m

Explanation / Answer

The law of conservation of mechanical energy states that the total mechanical energy of a closed system is constant. That means that the energy E1 at the time of releasing the block is equal to the energy E2 at the time the block stops. (E1=E2=konst) Mechanical energy comprises of potential energy and kinetic energy. Kinetic energy is (1/2)mv^2 and the potential energy is Ep=mgh+1/2 kx^2 (the first part being the PE of the gravitational field and the other one the PE of the spring compressed by some distance x) E=Ek+Ep=(1/2)mv^2+mgh+1/2 kx^2 but from the energy conservation law you know that: E1=E2 (1/2)m(v1)^2+mg(h1)+ +(1/2)k(x1)^2= =(1/2)m(v2)^2+mg(h2)+ (1/2) k(x2)^2 and since x1 is zero (at first, the spring is not compressed) and v2 is zero (the block is at rest), we can write this as => (1/2)m(v1)^2+mg(h1)=mg(h2)+1/2 k(x2)^2 now just plug in your values (remember that h=s*sin ?)

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