Question

1. A body of mass 1 kg is kept at rest. A constant force of 6 N starts acting on
it. Find the time taken by the body to move through a distance of 12 m.

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Answer 1

Answer :-

Given :-

• Initial velocity = 0 m/s
• Mass = 1 kg
• Force = 6 N
• Distance = 12 m

To Find :-

• Time taken

Solution :-

To find time, we first need to find acceleration and then substitute the value of given in the second equation of motion -

➩ Force = Mass × Acceleration

➩ 6 = 1 × a

➩ a = 6 m/s²

• u = 0
• a = 6 m/s²
• s = 12 m

Substituting the value in 2nd equation of motion :-

➩ s = ut + 1/2 at²

➩ 12 = 0 + 1/2 × 6 × t²

➩ 12 = 3t²

➩ t² = 12/3

➩ t² = 4

➩ t = 2 sec

Time taken = 2 sec

Answer 2

Given :-

• Initial Velocity = 0 m/s
• Mass = 1 kg
• Force = 6 N
• Distance = 12 m

To find :-

• Find the Time Taken ?

Solution :-

• Find Time Taken, We first need to find acceleration and then substitute the value of given in the second equation of motion.

$\underline{\bf{\bigstar} \: \mathfrak{As \: we \: know \: that}}$

• $\boxed{\bf{Force~=~Mass~×~Acceleration}}$★

• $\large\dashrightarrow$6 = 1 × a
• $\large\dashrightarrow$ a = 6 m/s²

• $\large\mapsto$0
• $\large\mapsto$6 m/s²
• $\large\mapsto$12 m

★ Substituting the value,

• 2nd equation of motion.

:$\implies$s = $\large{\sf{ut \: + \: \frac{1}{2} \: at²}}$

$~~~~~$:$\implies$$\large{\sf{0 \: + \: \frac{1}{2} \: \times \: 6 \: \times t²}}$

$~~~~~~~~~~$:$\implies$12 = 3t²

$~~~~~~~~~~~~~~~$:$\implies$t² = $\large{\sf{\frac{12}{3}}}$

$~~~~~~~~~~~~~~~~~~~~$:$\implies$t² = ${\sf{\dfrac{\cancel{12}}{\cancel{3}}}}$

$~~~~~~~~~~~~~~~~~~~~~~~~~$:$\implies$ $\large{\underline{\boxed{\pink{\frak{t~=~2~sec}}}}}$★

$\large\dag$ Hence,

• Time Taken = $\large\underline{\rm{2~sec}}$$\large{\bf\green{✓}}$