How do you find frictional force with no mass given?

[KingPhysics]

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!

 

[ideasrule]

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.

 

[Senjai]

I'm also quite interested in this question, i come up with two unknowns and only one equation.

 

[Senjai]

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:

F_{engine} = F_{net} + F_F - F_x

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

After substitution:

F_{engine} = ma + \mu mg \cdot cos\Theta - mg \cdot sin\Theta

Then i took the mass as the common factor:

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

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

i end up with F_{engine} = 3.37 \cdot m 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?

 

[denverdoc]

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.