Question

2. The force of 200 N acts on a moving body with velocity 15 m/s and is brought to rest after 5s. Calculate the mass of the given body. Also calculate the distance covered by the object after the application of force. plz ans step by step

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Mass\:of\:body=\frac{200}{3}\:kg}}}$

$\green{\tt{\therefore{Distance\:travelled=37.5\:m}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies Force(F) = 200 \: N\\ \\ \tt: \implies Initial \: velocity(u) = 15 \: m/s \\ \\ \tt: \implies Time(t) = 5 \: sec \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Mass \: of \: body(M) = \\ \\ \tt: \implies Distance \: travelled(s) =?$

• According to given question :

$\tt \circ \: Final \: velocity(v) = 0 \: m/s \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies Force = Mass \times Acceleration \\ \\ \tt: \implies - 200 = Mass \times \frac{v - u}{t} \\ \\ \tt: \implies -200 = Mass \times \frac{0 -15 }{5} \\ \\ \tt: \implies Mass = \frac{ - 200}{ - 3} \\ \\ \green{\tt: \implies Mass = \frac{200}{3} \: kg} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {v}^{2} = {u}^{2} + 2as \\ \\ \tt: \implies {0}^{2} = {15}^{2} + 2 \times ( - 3) \times s \\ \\ \tt: \implies - 225 = - 6 \times s \\ \\ \tt: \implies s = \frac{ - 225}{ - 6} \\ \\ \green{\tt: \implies s = 37.5 \: m} \\ \\ \green{ \tt \therefore Mass \: of \: body \: is \: \frac{200}{3} \: kg} \\ \\ \green{\tt \therefore Distance \: travelled \:by \: body \: is \: 37.5 \: m}$