Tired superman

We know the following to be true with respect to the block:
$Fg_\parallel = mg \sin\theta$ and $Fg_\perp = mg\cos\theta$

With some simply geometry, we find that the inclination of the ramp to the vertical is equal to the angle that the force of gravity makes with the normal force.
We find that $F_\perp = mg\cos(30) = 5000\cos(30) = 4330 Newtons$
Knowing that $F_f = F_\perp \times \mu = 4330 \times 0.4 = 1732 Newtons$

Now, it seems tempting to find $Fg_\parallel$ but it is not actually needed.

Notice:
The Force Needed to Stop the Car _ = $Fg_\parallel - F_f$ because friction helps stop the car.
_ The Force Needed to Move the Car = $Fg_\parallel + F_f$ because friction stops the car from moving upwards.
The question asks for the difference, and after subtracting these two equations we find that the answer is:
$2 \times F_f = 2 \times 1732 = \boxed{3464 Newtons.}$

A force diagram helps to keep everything straight on this one.

$F_{stop} = F_g \sin \theta - F_F$

$F_{move} = F_g \sin \theta + F_F$

$F_F = \mu F_N = \mu F_g \cos \theta = 0.4 F_g \cos 30$

$|F_{move} - F_{stop}| = |2 F_F| = |2 * 0.4 *5000 \cos 30| = 3464$

$F_{\text{normal}} = 5000 \cdot \cos(30^{\circ}) = 2500\sqrt{3} N,$

so the friction is

$0.4 \cdot 2500\sqrt{3} = 1000\sqrt{3} N.$

The difference in force of stopping the car and pushing it back up is the friction: when stopping the car, the friction helps you, while when pushing the car up, the friction is working against you. So, the difference we're looking for is twice the friction, which is

$2 \cdot 1000\sqrt{3} \approx \boxed{3460}.$

Minimum force required to prevent car from sliding= $=mg(\sin \theta-\mu \cos \theta)$ $=5000(\frac{ 1 }{ 2 }-0.4*\frac{ \sqrt{3} }{ 2 })=2500-1000\sqrt{3}$ Minimum force required to just push it up the incline= $=mg(\sin \theta +\mu \cos \theta)$ $=2500+1000\sqrt{3}$ so the difference is coming out to be $2000\sqrt{3}=3464$

mass of the car comes out to be 500 kg taking G=10..Firstly we need to calculate the force which our superman is needed to exert on the car that and for that we will have to resolve the component towards superman and that is {mg \times (\sin \theta)} - (4/10) {mg \times (\cos \theta)} . lets now get the force which superman must apply it comes out to be
{mg \times (\sin \theta)} + (4/10)
{mg \times (\cos \theta)}. when we subtract we get 2(4/10)*{mg \times (\cos \theta)} putting values i get 3460.000

• a moment before the car sliding down:

$∑F = ma = 0$

$F_{1} + f - W sin 30^{0} = 0$

$F_{1} = W sin 30^{0} - µ W cos 30^{0}$

• a moment the car starts sliding up:

$∑F = ma = 0$

$F_{2} - f - W sin 30^{0} = 0$

$F_{2} = W sin 30^{0} + µ W cos 30^{0}$

then finally we'll have:

$F_{2} - F_{1} = 3464.102 N$

By using FBD we can find it initial force required will be equal to 5000 (sin30-.4 cos30) then the force required to push is 5000 (sin30+.4 cos30) thus difference will yield 3460N as the answer.

When superman stops the car, consider the free-body diagram of the car. We have 4 forces: weight of the car $W$ , normal contact force $N$ , friction (which acts up the slope) $f=\mu N$ , and superman's force $F_1$

Resolving the forces perpendicular to the slope would give us $N=W\cos30^\circ$ , therefore $f=\mu W\cos30^\circ$

Resolving the forces parallel to the slope would give us $F_1 = W\sin30^\circ-f$

Now, in the case when superman pushes the car with the force $F_2$ so that it starts to move upwards, the frictional force $f$ , which still has the same magnitude, now acts down the slope. Therefore, resolving the forces parallel to the slope gives us $F_2 = W\sin30^\circ+f$

The difference between $F_1$ and $F_2$ is thus equal to $2f = 2\mu W\cos30^\circ = 3460N$