The magnitude of the vector product of two vectors `veca= 2i+5j` and `vecb=3j+4k` is

Answer

A, B, C, D

Explanation

a. Total kinetic energy
`y=1/2mv^2+1/2Iomega^2`
therefore, according to the question
`mgH=1/2mv^2+1/2Iomega^2+mgR(1+sintheta)`
`rarr mgH-mgR(1+sintheta)`
`=1/2mv^2+1/2Iomega^2`
`rarr 1/2mv^2+1/2Iomega^2=mg(H-R-Rsintheta)`
b. To find the acceleratiohn components,
`rarr 1/2 mv^2+1/2Iomega^2=mg(H-R-Rsintheta)`
`rarr k7/10 mv^2=mg(H-R-Rsintheta)`
`rarr v^2=10/7g(H-R-Rsintheta)`
`rarr v^2=10/7g(H-R-Rsintheta)`.....i
`rarr v^2/R=10/7 (g[H-R)-Rsintheta])/R`
`rarr v^2/R=(10g[(H-R)-Rsintheta)])/R`
`-2v (dv)/(dt)=-(10/7)g R costheta(dtheta)/(dt)`
`rarr omegaR(dv)/(dt)=-(5/7)gRcostheta (dtheta)/(dt)`
`rarr (dv)/(dt)=-(5/7)gcostheta`
`rarr ` tangential acceleration
c. Normal force at `thea=0`
`(mv^2)/R=(70/100)xx(10/7)xx10{((0.6-0.1))/0.1}`
`=5N`
Friction force:
`=f=mg-ma`
`-f=m(g-a)`
`m(10-5/7x10)`
`=0.07(10-5/7xx10)`
`=1/100(70-50)=0.2N`