Question

c) 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 a distance of 12 m. orov​

Answer 1

Answer:

The time taken is t = 2s.

We have F = 6N, m = 1kg, s = 12m and u = 0m/s.

From F = ma, on substituting the given values for F and m, we get a = 6m/s².

Now, in the equation s = ut+at², we substitute for s, u and a, to get t² = 4 ⇒ t = 2s (-2 is neglected since time cannot be negative).

Explanation:

F = m x a

Given, F = 6N

therefore, 6N = 1kg x a

               a = 6/1 = 6 m/s2

2as = (v)2 - (u)2

Given, u = 0 and s = 12 m

therefore, 2 x 6 x 12 = (v)2 - (0)2

               12 x 12 = v2

                144 = v2

               So, v = root of 144

                            12

Now take, v = u + at

Given, u = 0 and a = 6m/s

Therefore, 12 = 0 + 6t

                So, t = 12/6

                     =  2/1

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Answer 2

Answer:

2 seconds

Explanation:

F = mass * acceleration

Given - Force is 6 N. Mass is 1 kg.

=> $6 = 1 * a$

=> 6 = a

Thus, the body’s acceleration is 6 $ms^{-2}$

Using the Second Equation of Motion,

$s = ut + \frac{1}{2} at^{2}$

Here,

• u = 0 (Initial Velocity)
• t (To be found)
• s = 12 metres (Distance Travelled)
• a = 6 $ms^{-2}$

So,

=> $12 = 0t + \frac{1}{2}*6*t^{2}$

=> $12 = 3t^{2}$

=> $\frac{12}{3} = t^{2}$

=> $4 = t^{2}$

=> 2 = t

Time taken to travel is 2 seconds.