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A 6 kg object is at rest. When 30 N force is applied on it, the distance travelled by the object in 4 sec will be:
1. 80 m
2. 60 m
3. 48 m
4. 40 m

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Correct Answer - Option 4 : 40 m

CONCEPT:

Equation of Kinematics:

  • These are the various relations between u, v, a, t, and s for the particle moving with uniform acceleration where the notations are used as:
  • Equations of motion can be written as

⇒ V = U + at

\(⇒ s =Ut+\frac{1}{2}{at^{2}}\)

⇒ V2 =U2+ 2as

where, U = Initial velocity, V = Final velocity, a = Acceleration, t = time, and s = Displacement/Distance covered

CALCULATION:

Given m = 6 kg, F = 30 N, U = 0 m/s, and t = 4 sec

According to Newton's second law of motion,

⇒ F = ma

So,

\(⇒ a=\frac{F}{m}\)

\(⇒ a=\frac{30}{6}\)

⇒ a = 5 m/s2

By the second equation of motion,

\(⇒ s =Ut+\frac{1}{2}{at^{2}}\)

\(⇒ s =(0\times4)+\left ( \frac{1}{2}\times5\times 4^{2} \right )\)

\(⇒ s =(0\times4)+\left ( \frac{1}{2}\times5\times 4^{2} \right )\)

⇒ s = 40 m

  • Hence, option 4 is correct.

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