Question

A force 0.5N acts upon a body at rest and it moves a distance of 1m in 4s find the mass of the body

Answer 1

Answer:

• Mass (M) of The Body is 4 Kg.

Given:

1. Force Acting (F) = 0.5 N.
2. Distance Covered (S) = 1 m.
3. Time Taken (t) = 4 Sec.

Explanation:

$\rule{300}{1.5}$

From Second Kinematic Equation.

$\bigstar \; {\large{\boxed{\tt S = ut = \dfrac{1}{2}at^2}}}$

$\bold{Here}\begin{cases}\text{S denotes Distance Travelled } \\ \text{u Denotes Initial Velocity} \\ \text{a Denotes acceleration} \\ \text{t Denotes Time taken}\end{cases}$

Now,

${\large{\boxed{\tt S = ut = \dfrac{1}{2}at^2}}}$

Substituting the values,

$\longmapsto{\large{\tt 1 = 0 \times t + \dfrac{1}{2} \times a \times (4)^2}}$

$\longmapsto{\large{\tt 1 = 0 + \dfrac{1}{2} \times a \times 16}}$

$\longmapsto{\large{\tt 1 = \dfrac{1}{2} \times a \times 16}}$

$\longmapsto{\large{\tt a = 1 \times \dfrac{2}{16}}}$

$\longmapsto{\large{\tt a = 1 \times \cancel{\dfrac{2}{16}}}}$

$\longmapsto{\large{\tt a = 1 \times \dfrac{1}{8}}}$

$\longmapsto{\large{\underline{\boxed{\tt a = \dfrac{1}{8} \; m/s^2}}}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Applying Newton's Second law of motion.

$\bigstar \; {\large{\boxed{\tt F = Ma}}}$

$\bold{Here}\begin{cases}\text{F Denotes Force Applied} \\ \text{M Denotes Mass} \\ \text{a Denotes acceleration}\end{cases}$

Now,

${\large{\boxed{\tt F = Ma}}}$

Substituting the values.

$\longmapsto{\large{\tt 0.5 = M \times \dfrac{1}{8}}}$

$\longmapsto{\large{\tt M = 0.5 \times 8}}$

$\longmapsto{\large{\tt M = \dfrac{5 \times 8}{10}}}$

$\longmapsto{\large{\tt M = \dfrac{40}{10}}}$

$\longmapsto{\large{\tt M = \cancel{\dfrac{40}{10}}}}$

$\longmapsto{\large{\underline{\boxed{\red{\tt M = 4 \; Kg}}}}}$

∴ Mass (M) of The Body is 4 Kg.

$\rule{300}{1.5}$