Question
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
(a) Ignore friction
(b) Assume the effective coefficient of friction retarding the car is 0.30
Answers
bobpursley
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
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
Jen
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.
bobpursley
I get for a) 1.050E3*9.8*4.6E2*sin17.5=6.97E5 joules
Jen
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?