A body of mass 0.5 kg is resting on a frictionless surface.When a forc

1.	A body of mass 0.5 kg is resting on a frictionless surface.When a force of 2000 dyne acts on it for 10 s, then calculate the distance traveled by it in 10 seconds.
Answer» Required answer :- DISTANCE travelled by the body in 10 seconds is 2 m EXPLANATION :- As per the given information we have, MASS of the body is (m) 0.5 kg The body is resting of FRICTIONLESS surface [u =0] FORCE acting is 2000 dynes Time TAKEN is 10 s We have to find the distance travelled From equations of motion, We know that , $\implies \: s = ut + \dfrac{1}{2} at {}^{2}$ where, s = distance covered u = initial velocity t = time taken a = acceleration So, we can apply this formula But we don't know the acceleration of the body . So,Lets calculate acceleration of body. We know that, $\implies \: f = ma$ F = force m = mass a = acceleration F = 2000 dynes Converting dynes - Newtons $\implies{F = 2000\times 10^{-5} N }$ $\implies{F = 2 \times 10^3× 10^{-5} N}$ $\implies{F = 2× 10^{-2}N}$ F = m × a $\implies 2 \times 10 {}^{ - 2} = 0.5 \times a$ $\implies \dfrac{2}{100} = \dfrac{5}{10} \times a$ $\implies \dfrac{2}{10} = 5a$ $\implies \dfrac{1}{5} = 5a$ $\implies 25a = 1$ $\implies a = \dfrac{1}{25} \: m/s {}^{2}$ Substituting these values in formula $\implies \: s = ut + \dfrac{1}{2} at {}^{2}$ $\implies \: s = 0(10) + \dfrac{1}{2} \times \dfrac{1}{25} \times (10) {}^{2}$ $\implies \: s = \dfrac{1}{50} \times 100$ $\implies \: s = 2m$ So, distance travelled in 10 seconds by the body is 2m