Question

B) a body of mass 500g is kept at rest. a constant force of 6 newton
starts acting on it. find the time taken by the body to move to a distance of
1.2 km.

Answer 1

Given :

  Mass , m = 500 g = 0.5 kg

   Initial velocity ,  u = 0 m/s

                  Force , F = 6 N

Distance , s = 1.2 km = 1200 m

Formula Applied :

$\bigstar\ \; \sf \pink{s=ut+\dfrac{1}{2}at^2}\\\\\bigstar\ \; \sf \green{F=ma}$

where ,

• s denotes distance
• u denotes initial velocity
• a denotes acceleration
• t denotes time
• F denotes force
• m denotes mass

Solution

• F = 6 N
• m = 0.5 kg

$\sf F=ma\\\\\to \sf 6=(0.5)a\\\\\to \sf \red{a=12\ m/s^2}$

• s = 1200 m
• u = 0 m/s
• a = 12 m/s²

$\sf s=ut+\dfrac{1}{2}at^2\\\\\to \sf 1200=(0)t+\dfrac{1}{2}(12)t^2\\\\\to \sf 1200=6t^2\\\\\to \sf t^2=200\\\\\to \sf \blue{t \cong 14.14\ s}$

Answer 2

Given :-

• Mass = 500g = 0.5kg.

• Force = 6N.

• Distance = 1.2km.

• Initial Velocity = 0m/s.

To Find :-

• Time taken by the body.

Solution :-

We know that,

✰ F = ma ✰

Put the values.

➠ 6 = 0.5 × a.

➠ a = 60/5.

➠ a = 12m/s²

Now, Apply Second Equation Of Kinematics.

❂ s = ut + 1/2 at² ❂

➜ 1200 = 0t + 1/2 × 12 × t²

➜ 1200 = 6t²

➜ t = √200 = 14.14 Approx.

Hence, Your Answer is 14.14 seconds.