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]

Solution

Physics sir · 3 steps

Friction is the only force slowing the block, so find its deceleration μ g and the mass drops out.

1. Deceleration. Kinetic friction is μ mg, so the deceleration is a = (μ mg)/m = μ g = 0.5 × 9.8 = 4.9 m s⁻². The 10 kg mass cancels.

   v² = u² - 2as ⇒ s = u²/(2μ g)

2. Put in numbers. The block stops, so v = 0. Then s = (9.8)²/(2 × 4.9) = 96.04/9.8 = 9.8 m.
3. So. The block slides 9.8 m before coming to rest, which is option (b).

The idea to keep. On a level rough surface, the deceleration is μ g whatever the mass. Stopping distance is then u²/(2μ g).

Answer from the NTA final answer key. Solution by PrepWiser.