A spring (k = 220 N/m) is fixed at the top of a frictionless plane inclined at a

Question

A spring (k = 220 N/m) is fixed at the top of a frictionless plane inclined at an angle ? = 40° (Figure 8-66). A 1.0 kg block is projected up the plane, from an initial position that is distance d = 0.60 m from the end of the relaxed spring, with an initial kinetic energy of 14 J.

(a) What is the kinetic energy of the block at the instant it has compressed the spring 0.20 m?
J

(b) With what kinetic energy must the block be projected up the plane (from its initial position) if it is to stop momentarily when it has compressed the spring by 0.40 m?


J

Explanation / Answer

      The spring constant of the spring, k = 220 N/m       The mass of the block, m = 1.0 kg       The initial seperation between block and spring, d = 0.60 m       The initial kinetic energy, (K.E)i = 14 J       The given angle, = 40o _____________________________________________________________ a)       The length compressed spring, x = 0.20 m       The total distance of the block, is              d+x = 0.20 m+0.60 m                     = 0.8 m       The y-component of the displacement of the block is                   h = (0.8 m)sin40o                      = 0.514 m       Using the law of conservation of energy,               (K.E)i+(P.E)i = (K.E)+mgh+(1/2)kx2              14 J + 0 = (K.E)+(1.0 kg)(9.8 m/s2)(0.514 m)+(1/2)(220 N/m)(0.20 m)2       By solving,       The kinetic energy of the block at the instant it has       compressed the spring 0.20 m is                 K.E = 4.56 J _______________________________________________________________________ _______________________________________________________________________ b)       The compressed length, x' = 0.40 m       Using the law of conservation of energy,          (K.E)i+(P.E)i = (K.E)+mgh+(1/2)kx2               (K.E)i + 0 = 0 +(1.0 kg)(9.8 m/s2)(0.6 m +0.4 m)sin40o+(1/2)(220 N/m)(0.40 m)2       Therefore, the kinetic energy of the block is               (K.E)i= 23.89 J                           = 23.9 J                         By solving,       The kinetic energy of the block at the instant it has       compressed the spring 0.20 m is                 K.E = 4.56 J _______________________________________________________________________ _______________________________________________________________________ b)       The compressed length, x' = 0.40 m       Using the law of conservation of energy,          (K.E)i+(P.E)i = (K.E)+mgh+(1/2)kx2               (K.E)i + 0 = 0 +(1.0 kg)(9.8 m/s2)(0.6 m +0.4 m)sin40o+(1/2)(220 N/m)(0.40 m)2       Therefore, the kinetic energy of the block is               (K.E)i= 23.89 J                           = 23.9 J       Therefore, the kinetic energy of the block is               (K.E)i= 23.89 J                           = 23.9 J                          

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