i need help in physic?

Question

1) a 215N box is placed on an inclined plane that makes a 35.0 degree angle with the horizontal. find the component of the weight force parallel to the plane's surface

2)you push a 325 N trunk up a 20.0 degree inclined plane at a constant velocity by exerting a 211 N force parallel to the plane's surface.
- what is the component of the trunk's weight parallel to the plane
-what is the sum if all forces parallel to the plane's surface?
-what is the magnitude and direction of the friction force?
-what is the coefficient of friction?

Answer 1

215*sin(35)=123 N

2)
parallel is
325*sin(20)
111.1 N
The sum of all forces parallel is 0 since the velocity is constant
friction is
211-325*sin(20)
100 N
the coefficient is found by
(211-325*sin(20))/(325*cos(20))
0.327
j

Answer 2

1) the component of weight parallel to a surface is mgsin(theta), in this case that is 215xsin(35)=123.3N

2) a) component of gravity parallel to plane is 325xsin(20)=111.1N

b) sum of all forces parallel to plane must be zero since the box is moving at constant velocity (that means no acceleration and that means no net force)

c) along the plane, all forces must sum to zero; there is 111 N of gravity down the plane, 211 N pushing up the plane, so there must be 100N of friction acting down the plane so that the vector sum of forces is zero.

d) frictional force = coeff friction x normal force

normal force = Wcos(theta) = 325cos(20)=305.4N

so coeff of friction = 100/305=0.33