A 3.0 kg object is released from rest at a height of 4.5 meters on a cursed, fri

Question

A 3.0 kg object is released from rest at a height of 4.5 meters on a cursed, frictionless ramp. At the foot of the ramp is a spring of force constant k = 100 N/m. The object slides down the ramp and into the spring, compressing it a distance Delta x before coming to rest. Use conservation of energy to solve the following: What is the gravitational potential energy of the object before it is released? What is the kinetic energy of the object at the bottom of the ramp ? What is the speed v of the object at the bottom of the ramp? How much is the spring compressed by the object? A 40.0-cm-radius wheel is rotating at a constant 5.0 rev/s. What is the period of rotation T? What is the speed v of a point on the outer edge of the wheel? What is the radial acceleration a_c of the point on the outer edge of the w heel ?

Explanation / Answer

9. Mass of the object M = 3 kg

Height H= 4.5 m

Spring constant K = 100 N/m

Energy Conservation

(a.) Gravitational potential energy = MgH = 3*9.8*4.5 = 132.3 joule

(b.) Since the energy will be conserved hence the initial potential energy will get converted to kinetic energy

Kinetic Energy at bottom of ramp = 132.3 joule.

(c.)   Kinetic Energy at bottom of ramp = 132.3 joule

or 1/2 MV2 = 132.3

or V = 9.39 m/s

(d.) 1/2 KX2 = 132.3

or X = 1.627 m

Hence compression in spring = 1.627 m

10.)

Radius R= 40 cm = 0.40 m

Angular speed = 5 rev / sec = 5*2*3.14 rad/sec = 31.4 rad/ sec

(a.) Period of rotation = 1/5 seconds = 0.2 sec

(b.) Speed V = Angular speed *R = 31.4*0.40 = 12.56 m/s

(c.) Radial acceleration = V2/R = 394.38 m/s2

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