Question

 A block of mass 10 kg starts sliding on a surface with an initial velocity of 9.8 ms^–1. The coefficient of friction between the surface and block is 0.5. The distance covered by the block before coming to rest is[use g = 9.8 ms^–2]

• A. 4.9 m
• B. 9.8 m
• C. 12.5 m
• D. 19.6 m

Answer 1

The Correct Option is B

 To solve this problem, we need to find the distance covered by the block before it comes to rest using the given parameters: mass of the block \(m = 10 \, \text{kg}\), initial velocity \(u = 9.8 \, \text{ms}^{-1}\), coefficient of friction \(\mu = 0.5\), and acceleration due to gravity \(g = 9.8 \, \text{ms}^{-2}\).

The force of friction \((f)\span> acting on the block can be calculated using the formula:\)

\(f = \mu \cdot N\),

where \(N\) is the normal force, which equals the weight of the block since the block is sliding on a horizontal surface. Thus, \(N = m \cdot g = 10 \cdot 9.8 = 98 \, \text{N}\).

Substituting the values, we get:

\(f = 0.5 \times 98 = 49 \, \text{N}\).

This frictional force causes a deceleration \((a)\span>, which can be found using Newton’s second law:\)

\(f = m \cdot a\)

\(49 = 10 \cdot a\)

\(a = 4.9 \, \text{ms}^{-2}\) (negative because it is deceleration).

We use the kinematic equation to find the distance \((s)\) the block travels before coming to rest:

\(v^2 = u^2 + 2as\).

Given that the final velocity \(v = 0 \, \text{ms}^{-1}\) when the block comes to rest, we have:

\(0 = (9.8)^2 + 2(-4.9)s\).

We solve for \(s\):

\(0 = 96.04 - 9.8s\)

\(9.8s = 96.04\)

\(s = \frac{96.04}{9.8} = 9.8 \, \text{m}\).

Therefore, the block covers a distance of 9.8 meters before coming to rest. Thus, the correct answer is 9.8 m.