# What is the minimum work needed to push a 1000 kg car 490 m up along a 17.5° incline? Ignore friction. Ignore friction. Assume the effective coefficient of friction retarding the car...

What is the minimum work needed to push a 1000 kg car 490 m up along a 17.5° incline? Ignore friction.

Ignore friction.

Assume the effective coefficient of friction retarding the car is 0.35?

*i've tried 8575 for ignore friction and got it wrong. can someone please help with this and list the steps? i get confused around the part where you have to plug in cos and sin.*

### 1 Answer | Add Yours

For a diagram of the forces please see the figure below.

The weight `G` need to be decomposed into two: the normal to the inclined plane component `G_n` and the parallel to the inclined plane component `G_p` . The angle `alpha` of the inclined is also found between `G` and `G_p`. Because of this

`G_n = G*cos(alpha)`

`G_p =G*sin(alpha)`

If we ingnore friction the only force acting along the inclined plane is `G_p` and the work done by this force along 490 m is

`W_1 =G_p*d =m*g*d*sin(alpha) =1000*9.81*490*sin(17.5) =`

`=1445462 J =1.44*10^6 J =1.44 MJ`

If there is friction, the force acting along the plane is simply the sum of `G_p` and friction force `F_f` :

`F =F_f + G_p = mu*G_n +G_p =mu*G*cos(alpha) +G*sin(alpha)=`

`=0.35*1000*9.81*cos(17.5) +1000*9.81*sin(17.5) =6224.5 N`

` ` The work done by this force along the incline over a distance `d =490 m` is

`W_2 = F*d = 6224.5*490 =3050010 J =3.05*10^6 J =3.05 MJ`

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