A body is coming with a velocity of 72 km/h on a rough horizontal surface of coefficient of friction 0.5. If the acceleration due to gravity is 10 m/s2, find the minimum distance it can be stopped.

Question

A body is coming with a velocity of 72 km/h on a rough horizontal surface of coefficient of friction 0.5. If the acceleration due to gravity is 10 m/s2, find the minimum distance it can be stopped. (A) 400 m (B) 40 m (C) 0.40 m (D) 4 m

Tardigrade Admin · Accepted Answer

Given, u=72 km / h =20 m / s a=μ g=0.5 × 10 m / s 2 From v2=u2-2 as ∴ (0)2=(20)2-2 × 0.5 × 10 × s ∴ s=(20 × 20/2 × 0.5 × 10) or s=40 m

Q. A body is coming with a velocity of $72 km / h$ on a rough horizontal surface of coefficient of friction $0.5.$ If the acceleration due to gravity is $10 m / s^{2},$ find the minimum distance it can be stopped.

400 m

40 m

0.40 m

4 m

Given, $u = 72 km / h = 20 m / s$
$a = μg = 0.5 \times 10 m / s^{2}$
From $v^{2} = u^{2} - 2 a s$
$∴ (0)^{2} = (20)^{2} - 2 \times 0.5 \times 10 \times s$
$∴ s = \frac{20 \times 20}{2 \times 0.5 \times 10}$
or $s = 40 m$

Q. A body is coming with a velocity of 72km/h on a rough horizontal surface of coefficient of friction 0.5. If the acceleration due to gravity is 10m/s2, find the minimum distance it can be stopped.

400 m

40 m

0.40 m

4 m

Given, u=72km/h=20m/s a=μg=0.5×10m/s2 From v2=u2−2as ∴(0)2=(20)2−2×0.5×10×s ∴s=2×0.5×1020×20​ or s=40m

Q. A body is coming with a velocity of $72 km / h$ on a rough horizontal surface of coefficient of friction $0.5.$ If the acceleration due to gravity is $10 m / s^{2},$ find the minimum distance it can be stopped.

Given, $u = 72 km / h = 20 m / s$
$a = μg = 0.5 \times 10 m / s^{2}$
From $v^{2} = u^{2} - 2 a s$
$∴ (0)^{2} = (20)^{2} - 2 \times 0.5 \times 10 \times s$
$∴ s = \frac{20 \times 20}{2 \times 0.5 \times 10}$
or $s = 40 m$