Question

A brick is given an initial speed of 5 m/s up an inclined plane at an angle of 30⁰ from the
horizontal. The coefficient of (sliding or static) friction is = √3/12 . After 1.5 s, how far
is the brick from its original position? Assume g = 10 m/s2
.

Answer 1

Answer:

S= 0.47 m  

Explanation:

The retarding force actine on the block is given by;

Fp = mg sin θ+ μk(mg)cosθ

Fp = mg sin 30+ μk(mg)cos30

μk = √3/12 = 0.144

Fp = m (10) sin 30 + 0.14 (m)(10) cos 30

Fp = 5m + 1.25m

Fp= 6.25 m

Fp = ma

6.25m = m a

A= - 6.25 m/s^2  ( Negative as block is slowing down, moving against gravity)

S = ut + ½ at^2

T = 1.5

U = 5

Putting all the values;

S = ut + ½ at^2

S = 5 (1.5) + ½  (-6.25) (1.5)^2

S = 7.5 - 7.03

S= 0.47 m