A spring ( k = 195 N/m) is fixed at the top of a frictionless plane inclined at

Question

A spring (k = 195 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 20 J.

A spring (k = 195 N/m) is fixed at the top of a frictionless plane inclined at an angle ? = 40degree(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 20 J. What is the kinetic energy of the block at the instant it has compressed the spring 0.20 m? 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?

Explanation / Answer

a)Initial KE = final KE + PE of spring + PE of mass
=>20 = final KE + 0.5*195*0.2^2 + 1*9.8*0.8*sin40*
=>final KE = 11.06 J

b)Initial KE = final KE + PE of spring + PE of mass
=>Initial KE= 0 + 0.5*195*0.4^2 + 1*9.8*1.0*sin40*
=>Initial KE = 21.9 J

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