Question

17.
A block with initial velocity 4.0 m/s slides 8.0 m across a rough horizontal floor before coming
to rat. The coefficient of friction is:
1) 0.80
2) 0.40
3) 0.20
4) 0.10​

Answer 1

Answer: option (4): 0.10

Explanation:

The initial velocity of the block, u = 4 m/s  

Distance travelled by the block across the horizontal floor, s = 8 m

Since the block is coming to rest, so the final velocity, v = 0 m/s

Based on the eq. of kinematics, we have

v² = u² + 2as

⇒ 0 = (4)² + 2*a*8 ….[substituting the given values]

⇒ -16a = 16

⇒ a = - 1 m/s² ← -ve sign denotes the deaccelerated motion of the block, so, the magnitude of the acceleration is 1m/s²

Now, by using Newtons’ second law of motion,

Frictional force, F = m*a ….. (i)

But, the frictional force is also given as,

F = µk * m * g …… (ii) …. [here µk = coefficient of kinetic friction]  

Comparing eq. (i) & (ii), we get

m * a = µk * m * g

substituting a = 1 m/s² & g = 9.8 m/s²

⇒ µk = 1/9.8 = 0.10

Thus, the coefficient of friction is 0.10.