Question

Question #39b14

Answer 1

(a) Given street incline `= 20.0^@`
Let Distance moved by Chemistro along the street`=l`
`:.` the vertical height reached by Chemistro `h=lsin20^@`

Total energy of Chemistro at any instant of time `=PE+KE`

`=mgh+1/2mv^2`
Where `m` is mass of Chemistro and `g=10ms^-2` acceleration due to gravity

Initial total energy `=0+1/2xx90.0xx(22.0)^2=21780J`
Initial kinetic energy gets converted in `PE` of reaching at height `h` and rest is spent in doing work done against friction.

Final total energy when he comes to rest `=90.0xx10xxlsin20` `=900lsin 20^@J`

Now work done by friction
`W = "Force of Friction"xx"Distance moved up"=vec(F_f)cdot vecd`

From the figure we see that the normal reaction

`N = mgcos20^@`

therefore force of the sliding friction is

`F_f= μ_kN = μ_kmgcos20^@`
`:.W=-μ_kmgcos20^@xxl`
As force of friction and distance moved are opposite to each other we have `-ve` sign.

Inserting given values and using law of conservation of energy we get

`21780=900lsin 20^@+0.333xx90.0xx10lcos20^@`
`=>21780=900lxx0.3420+0.333xx90.0xx10lxx0.9397`
`=>21780=(307.8+281.6)l`
`=>l=37.0m`, three significant figures
`:.h=36.95xxsin20^@`
`=>h=12.6m`

(b) Heat generated by sliding is same as work done against friction

`=0.333xx90.0xx10xxlcos20^@`
`=0.333xx90.0xx10xx36.95xx0.9397=10406J`

(c) We know that power is the rate of doing work per unit time.

`:.` Average power delivered by Power Man `=21780/0.300` `=72600Js^-1`