How do you find frictional force with no mass given?

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KingPhysics
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Homework Statement


Batman is driving the Batmobile down a hill coming from the Bat Cave. This hill is inclined at an angle of 30 degrees to the horizontal and has a coefficient of kinetic friction of 0.28. What force must the Batmobile's engine apply to cause the Batmobile to accelerate at 0.60g?

a = 0.60g = 0.60(9.8) = 5.88
uk = 0.28


Homework Equations


Fengine = ma + mg sin30 + Ff


The Attempt at a Solution


I tried plugging in what I had into the above equation, assuming that the mass was 0.60, so I got Fengine = (0.6)(5.88) + (0.6)(9.8)sin30 + 1.64 and my answer was 8.1 N as the force applied on the engine. But when I looked at the back of the book, it said that the answer is (3.36)(mass) N. How does that work? I am so confused!
 
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0.6 kg is about the weight of five apples, whereas the Batmobile is the size of a tank. How can you assume that m is 0.6 kg? Just call it "m" and assume it's known.

First, draw a free-body diagram of the Batmobile. Remember to label all forces. Then write Newton's second law for the direction parallel to the ramp and the direction perpendicular to it.
 
I'm also quite interested in this question, i come up with two unknowns and only one equation.
 
I attempted this question, i myself am reviewing for my standardized year-end tests..

though I am not sure it is correct.

i stated that:

[tex]F_{engine} = F_{net} + F_F - F_x[/tex]

Where Fx and Fy are the components of the gravitational force on the object.

After substitution:

[tex]F_{engine} = ma + \mu mg \cdot cos\Theta - mg \cdot sin\Theta[/tex]

Then i took the mass as the common factor:

[tex]F_{engine} = m\left( a + \mu g \cdot cos\Theta - g \cdot sin\Theta\right)[/tex]

And then i solve:
[tex]F_{engine} = m\left( (0.60)(9.8) + (0.28)(9.8)(cos(30)) - (9.8)(sin(30))\right)[/tex]

i end up with [tex]F_{engine} = 3.37 \cdot m[/tex] Newtons

which is as much as i could simplify, i don't think you can fully solve this question without the mass.. am i correct?
 
yes, unless the acceleration could be provided by gravity itself:

i.e sin(a) - u cos(a) = 0.6 Your answer is in agreemnet with that in the book according to the OP.