A block of mass 4.5 kg is sitting on a frictionless ramp with a spring at the bo

Question

A block of mass 4.5 kg is sitting on a frictionless ramp with a spring at the bottom, as shown, that has a spring constant of 515 N/m. The angle of ramp with respect to the horizontal is 18 A block of mass 4.5 kg is sitting on a frictionless ramp with a spring at the bottom, as shown, that has a spring constant of 515 N/m. The angle of ramp with respect to the horizontal is 18 A block of mass 4.5 kg is sitting on a frictionless ramp with a spring at the bottom, as shown, that has a spring constant of 515 N/m. The angle of ramp with respect to the horizontal is 18 Degree. Randomized Variables m = 4.5 kg k = 515 N/m ? = 18 Degree Part (a) The block, starting at rest, slides down the ramp a distance 38 cm before hitting the spring. How far, in centimeters, does the spring compress before the block stops? Part (b) After the block comes to rest, the spring pushes the block back up the ramp. How fast, in meters per second, is the block moving right after it comes off the spring? Part (c) What is the difference in gravitational potential energy between the original position of the block at the top of the ramp and the position of the block when the spring is fully compressed?

Explanation / Answer

apply EPE = GPE

0.5 kx^2 = mgh sin theta

where k is spring constnt

x is compression

m is mass

g is accleratrion due to gravity

h = d = 0.38 m

so x^2 = 4.5 * 9.8 * 0.38 * sin 18/(0.5 * 4.5)

x^2 = 2.301

x = 1.517 m----------------<<<<<<<<Answer

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apply KE = EPE

0.5 mv^2 = 0.5 kx^2

v^2 = 515 * 1.517 *1.517/(4.5)

v^2 = 263.369

v = 16.22 m/s

-----------------------------------

chnage of energy U = 0.5 mv^2 - mgh

U = (0.5* 4.5 * 16.22*16.22) - (4.5 * 9.8 * 0.32)

U = 577.83 Joules

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