A loaded truck of mass 3000 kg moves on a level road at a constant speed of 6.00 0 m/s. The frictional force on the truck from the road is 1000 N. Assume that air drag is negligible.
(a) How much work is done by the truck engine in 10.00 min ?

Explanation

KEY IDEA
When the truck starts to climb the hill, the power needed to overcome the gravitational pull of the Earth is
`P_(1) = F_(g) v = ( m g sin theta)v`
` = (3000 kg ) (9.8 m//s^(2)) sin 30^(@) (6.000 m//s)8.820 xx 10^(4) W`
Calculations: Similarly, the power needed to overcome friction is
`P_(2) = (mu mg cos theta) v = ( f cos theta) v`
`= (1000 N) cos 30^(@) (6.000 m//s) = 5.196xx10^(3) W`
Therefore, the total work done in 10 minutes is
`W. = (P_(1)+P_(2))t`
` = (8.820 xx 10 W + 5.196 xx 10^(3) W) ` (10 min) (60 s/min)
`= 5.604 xx 10J`
or about 56 MJ.