A 500 g solid disk roles down an incline plane 1 m long elevated by 10 cm. What is the kinetic energy of the disk at the bottom of the plane?
The radius of the disk is 10 cm. How fast is the disk moving, in m/s, at the bottom of the plane?
Using the law of conservation of energy yields that potential energy(PE) is equal to kinetic energy(KE) such that:
`PE = m*g*h`
The problem provides the mass of `500g = 0.5Kg ` and the elevation of `10cm = 0.1m` , hence, substituting these values in formula of PE yields:
`PE = 0.5*9.8*0.1 = 0.49 m^2*kg/s^2`
You should remember that the total kinetic energy of the disk is the sum of kinetic energy of disk and kinetic energy of the falling mass such that:
`KE = I(omega)^2/2 + mv^2/2`
You need to find the moment of inertia of the disk such that:
`I = m*R^2/4`
You need to find the angular speed omega such that:
`omega = v/R`
`KE =m*R^2*v^2/4*R^2 + mv^2/2`
`KE = mv^2/4 + mv^2/2`
`KE = 3mv^2/4`
By the law of conservation of energy yields:
`KE = PE =gt 3mv^2/4 =m*g*h =gt 3v^2 = 4gh =gt v^2 = 4gh/3`
`v = 2sqrt(gh/3)`
Substituting 9.8 for g, 0.1 for h yields:
`v = 2sqrt(9.8*0.1/3) =gt v = 1.143 m/s`
You may evaluate the kinetic energy such tht:
`KE = 3mv^2/4 =gt KE =(3/4)*0.5*(1.143)^2 = 0.489 J`
Hence, evaluating the kinetic energy of the disk at the bottom of the plane yields `KE = 3mv^2/4 =0.489 J` and the speed`v = 1.143 m/s` .