accurate to within 2%. 63 An inclined plane of 20.00 has a angle spring of force

Question

accurate to within 2%. 63 An inclined plane of 20.00 has a angle spring of force constant k 500 N/m fastened securely at the bottom so that the spring is parallel to the surface as shown in Figure P7.63. A block Figure P7.63 of mass m 2.50 kg is Problems 63 and 64. placed on the plane at a distance d 0.300 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?

Explanation / Answer

Let 'x' be the compression in the spring when the block comes to rest

Let the initial position of the block represent zero potential energy

Conserving energy,

KEi + PEi + SPEi = KEf + PEf + SPEf (KE= kinetic energy, PE= Potential energy, SPE= Spring potential energy)

=> (1/2)mv2 + 0 + 0 = 0 - m*g*(d+x)*sin + (1/2)kx2

=> kx2 - (2*m*g*sin)x - m(v2 + 2*g*d*sin) = 0

Using the formula for solution of a quadratic equation,

=> x = [2*m*g*sin + (4*m2*g2*sin2 - 4*k*(-m(v2 + 2*g*d*sin)))]/2k

=> x = [2*m*g*sin + (4*m2*g2*sin2 + 4*k*(m(v2 + 2*g*d*sin)))]/2k

=> x = [m*g*sin + (m2*g2*sin2 + k*(m(v2 + 2*g*d*sin)))]/k

=> x =  [m*g*sin + (m2*g2*sin2 + kmv2 + 2*k*g*d*sin)]/k

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