Question

Particle of mass 0.1 kg has an initial speed of 4 ms-1 at a point a on a rough horizontal road. The coefficient of friction between the object and the road is 0.15. The particle moves to a point b, at a distance of 2 m from



a. What is the speed of the particle at b ? (take g

Answer 1

initially, particle placed at point a, initial speed of particle, u = 4 m/s

particle moves to point b, at a distance of 2m from a.

so, distance travelled by particle, s = 2m

mass of particle , m = 0.1 kg

according to question, surface is rough so, frictional force acts between particle and surface.

e.g., frictional force = ma

or, $\mu mg$ = ma

or, $a=\mu g$ = 0.15 × 10 = 1.5m/s²

hence, retardation acts on particle, a = -1

5 m/s²

now use formula, v² = u² + 2as

or, v² = 4² - 2 × 1.5 × 2 = 16 - 6

or, v² = 10 => v = √10

hence, speed of the particle at point b is √10m/s

Answer 2

$\bf\huge\textbf{\underline{\underline{According\:to\:the\:Question}}}$  

u = 4 m/s

Distance travelled by particle = 2m (s)

m = 0.1 kg

Using Formula we get :-

F = ma

$\bf\huge\bf\huge{\boxed{\bigstar{{a =\mu g}}}}$        

⇒ 0.15 × 10 = 1.5m/s²

Retardation (a) = -1.5 m/s²

$\bf\huge\bf\huge{\boxed{\bigstar{{Using \:Third\: Equation\: of\: motion}}}}$

v² = u² + 2as

v² ⇒ 4² - 2 × 1.5 × 2

v² = 16 - 6

v² ⇒ 10

v = √10

$\bf\huge\bf\huge{\boxed{\bigstar{{Speed \:of \:the \:particle\: is\: \sqrt{10}m/s}}}}$

$\bf\huge\bf\huge{\boxed{\bigstar{{Or\:=\:3.162\:m/s}}}}$