Question

a force of 10 n is aplied for 0.1 sec on a body of mass 1 kg initially at rest. what will be the velocity of the body just after this? After two more seconds with what velocity the body will move​

Answer 1

Answer:

Velocityafter2sec=21m/s

\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}Step−by−stepexplanation:

\begin{gathered} \green{\underline \bold{Given :}} \\ \tt: \implies Force(F) = 10 \: N \\ \\ \tt: \implies Mass(M) = 1 \: kg \\\\ \tt:\implies Time(t_{1})=0.1\:sec\\ \\ \tt: \implies Time(t_{2}) = 2 \: sec \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies velocity \: at \: 2 \: sec = ?\end{gathered}Given::⟹Force(F)=10N:⟹Mass(M)=1kg:⟹Time(t1)=0.1sec:⟹Time(t2)=2secToFind::⟹velocityat2sec=?

• According to given question :

\begin{gathered} \bold{As \: we \: know \: that} \\ \tt: \implies F = Ma \\ \\ \tt: \implies 10 = 1 \times a \\ \\ \green{\tt: \implies a = 10 { \: m/s}^{2} } \\ \\ \tt \circ \: Initial \: velocity = 0 \: m/s \\ \\ \tt \circ \: Time = 0.1 \: sec \\ \\ \tt \circ \: Acceleration = 10 { \: m/s}^{2} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies v = u + at \\ \\ \tt: \implies v = 0 + 10 \times 0.1 \\ \\ \tt: \implies v = 0 + 1 \\ \\ \green{\tt: \implies v = 1 { \: m/s}} \\ \\ \tt \circ \: Initial \: velocity = 1 \: m/s \\ \\ \tt \circ \: Time = 2 \: sec \\ \\ \tt \circ \: Acceleration = 10 { \: m/s}^{2} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies v = u + at \\ \\ \tt: \implies v = 1 + 10 \times 2 \\ \\ \tt: \implies v =1 + 20 \\ \\ \green{\tt: \implies v = 21 { \: m/s}}\\\\ \green{\tt \therefore Velocity \: after \: 2 \: sec \: is \: 21 \: m/s}\end{gathered}Asweknowthat:⟹F=Ma:⟹10=1×a:⟹a=10m/s2∘Initialvelocity=0m/s∘Time=0.1sec∘Acceleration=10m/s2Asweknowthat:⟹v=u+at:⟹v=0+10×0.1:⟹v=0+1:⟹v=1m/s∘Initialvelocity=1m/s∘Time=2sec∘Acceleration=10m/s2Asweknowthat:⟹v=u+at:⟹v=1+10×2:⟹v=1+20:⟹v=21m/s∴Velocityafter2secis21m/s