(Physics 12 help) Another Dynamics question on an object sliding down an incline...?

Question

The mass of a box is 50kg sliding down a bank angled 30 degrees to the horizontal. The coefficient of friction is 0.20.(a)
What is the acceleration of the box? (b) What is the frictional force on the box?

Answer 1

Draw a diagram of the box sliding on the incline and you see that the weight ( acting vertically ) of 50Kg can be resolved into a component parallel to the incline and one perpendicular to it. The former is 50Kg (sin 30)= 50Kg (1/2); the latter is 50Kg cos30 = 50 (sqrt 3)/2 . The latter acts opposite to the former through the frictional force which is along the incline, decelerating the box and with a force = 25(sqrt 3) X (0.2). This is the frictional force ( always decelerating ) on the box so th acceleration is
50Kg X a = 25 Kg - 5Kg(sqrt 3_). The acceleration is a =
0.5 - 0.1(sqrt 3) = 0.33 m/sec**2

Answer 2

Component of wt acting parallel to the inclined plane= mg sin theta
Component of wt acting parallel to the inclined plane=50 sin 30

Component of wt acting parallel to the inclined plane=25 N

Normal reaction of inclined plane R=mg cos theta

Normal reaction of inclined plane R=50cos30

Normal reaction of inclined plane R=50x0.867=43.35 N

Coefficient of friction (mu)=0.20

friction force 'f'= ( mu ) R=0.2 x43.35=8.67 N

The component of weight is more than friction and these are in opposite direction, their resultant R is equal to their difference.
Resultant R=25 - 8.67=16.33 N

Acceleration 'a'=resultant R/ mass

Acceleration 'a'=16.33 / 50 =0.3266 m/s^2

(a) Acceleration of the box is o.3266 m/s2 parallel to the plane downwards'

(b( frictional force is 8.67 N

Answer 3

Uhhhhhhhh...................I only in the 7 grade and they haven't taught me this.So I sorry i can not answer this.

Answer 4

3.14 = pi