Question

Friction: A block of mass 2kg is gently placed over a massive plank moving horizontally over a smooth surface with velocity 6m/s. The coefficient of friction between the block and plank is 0.2. Prove that the distance travelled by the block with respect to plank till it slides on the plank is 9m. (Take g =10 m/s^2)

40 POINTS. WILL MARK BRAINLIEST FOR PROPER ANSWER.
TROLL ANSWERS WILL BE DELETED

THANKS A LOT​

Answer 1

mass of block , m = 2kg

coefficient of friction between block and mass , u = 0.2

velocity of block, v = 6m/s

due to frictional force acting on block, its speed gradually decreases and finally block comes to rest. let after taking c distance block comes to rest.

from work energy theorem,

or, change in kinetic energy = workdone due to frictional force

or, (Kf - Ki) = Fr. x cos180° [ actually friction force acts just opposite of relative motion. so, angle between x and friction force is 180°]

final kinetic energy , Kf = 0 [ as finally block comes to rest.]

initial kinetic energy, Ki = 1/2 mv²

and friction force = umg

so, 0 - 1/2 mv² = umg × x cos180°

or, -1/2 v² = ug × x [-1]

or, v²/ug = x

or, x = (6)²/(0.2 × 2 × 10)

= 36/4 = 9

hence, distance covered by the block is 9m

Answer 2

Explanation:

A mass of recently accumulated to partially carbonized plant debris. Peat is an organic sediment. Burial, compaction, and coalification will transform it into coal, a rock. It has a carbon content of less than 60% on a dry ash-free basis.