physics question help help?

Decompose the weight force, that is always vertical,

in two components, one along the plane and the other

normal to the plane.

The first one has the down hill direction along the

plane. The second one will be responsable of the friction,

I remember to you that the friction acts on the conctact

interface and is always against to the motion.

Fw = M*g = 10*9.81 = 98.1 N --> kg*m/s^-2

Fn = Fw * cos(30) = 98.1 * 0.866 = 84.957 N

Fp = Fw * sin(30) = 98.1 * 0.5 = 49.05 N

Now I use kinetic friction to compute the friction

resistance

Fr = Fn * 0,4 = 84.957 * 0.4 = 33.983 N

Now I compute all forces acting on the mass in dynamic

situation, taking care to the direction of them.

Acting to rise the mass position (up hill)

Pulling force = 120N

Acting in opposite direction

Fp = 49.05 N

Fr = 33.983 N

Fa = m * a <-- accelerating force (if the mass

is able to move in uphill direction)

120 = 49.05 + 33.983 + Fa --> Fa= 120-83 = 37N

Fa is positive, then the mass will be able to move and

accelerate while pulled up, the acceleration will be:

a = Fa/m = 37/10 = 3.7 m/s^-2

The motion law for the constant acceleration starting still is:

S = 1/2 a t^2

t = sqrt(2*S/a) --> time after 5m of travel (t@5m)

t = sqrt(10/3.7) = 1,644 s

The speed of the mass increase linearly with the time,

and it is stopped at the start time t=0:

v(t)= a t

v(t@5m) = 3.7 * 1,644 = 6,08 m/s (ask value)

Bye

First find the resolved components of the weight and pulling force 'normal' to the plane, so that the friction force can be calculated:

Weight resolved = 10*9.81*cos30

Pulling force resolved = 120*sin30

Friction force down the plane = 0.4*[10*9.81*cos30+120*sin30]=56.75 N

Pulling force resolved 'up' the plane = 120*cos30=106.92 N

Net accelerating force 'up' the plane = 106.92-56.75=50.17 N

Acceleration = F/m=50.17/10=5.017 m/s^2

Use v^2=u^2+2as to determine the final velocity 'v':

v^2=0+2*5.017*5

v=7.08 m/s