A4000 kg truck is parked on a 7.0° slope. how big is the friction force on the truck?

4777.28 N

Step-by-step explanation:

The gravitational force down the slope is given by the formula;

(mass *g * sin θ )  

where;

θ is the angle of inclination of the slope

g is the gravitational pull

We plug in the given values into the above expression;

(4,000 x 9.8 x sin 7) = 4777.28 N.  

The frictional force must equal to 4777.28 N (pointing up the slope).

c

step-by-step explanation: