work= weight*height= m*g* 460SinTheta.
Now with friction, add to the basic work in a), the work done on friction, which is normal component of weight*coefficent*distance
or
friction= mg*CosTheta*mu*460
What is the minimum work needed to push a 1050 kg car 460 m up along a 17.5° incline?
(a) Ignore friction
(b) Assume the effective coefficient of friction retarding the car is 0.30
4 answers
I took m*g*460 sintheta
I used 1050*9.8* 460sin 17.5 and that's not the right answer I think I messed up somewhere.
I used 1050*9.8* 460sin 17.5 and that's not the right answer I think I messed up somewhere.
I get for a) 1.050E3*9.8*4.6E2*sin17.5=6.97E5 joules
okay i got 14.2e5 for a which is right but for b I keep getting it wrong, i used mg*CosTheta*mu*460
and I used 1.050e3*9.8*cos17.5*.30*460
but that aswer is not right I am not sure is the friction .30 =mu?
and I used 1.050e3*9.8*cos17.5*.30*460
but that aswer is not right I am not sure is the friction .30 =mu?
1 answer