Question

a body of a mass 2 it is moving with a velocity 8 m/s on a smooth surface. if it is to be brought to the rest in 4 sec then the force to be applied is​

Answer 1

Answer:

f = ma

f = m * v-u/t

= 2 * 0 - 8/4

= 2* -2

= -4

hope it helps you

Answer 2

Explanation:

$\huge \bold{Answer - }$

$\bold{Given - } \\ \bold{u = 8 \: ms {}^{ - 1} \: \:(initial \: velocity) }</p><p> \\ \bold{v = 0 \: ms {}^{ - 1} \: \: (final \: velocity) } \\ \bold{t = 4 \: s} \\ \\ \fbox\bold{s \: = (\frac{v + u}{2} )t} \\ \\ \leadsto \: ( \frac{0 + 8}{2} )4 \\\leadsto 4 \times 4 \\ \leadsto 16 \:m \\ \fbox \bold{s \: = 16 \: m} \\ \\ \\ \fbox\bold{s = ut + \frac{1}{2} at {}^{2} } \\ \leadsto16 = 8 \times 4 + \frac{1}{2} \times a \times (4) {}^{2} </p><p> \\ \leadsto16 = 32 + \frac{16a}{2} \\ \leadsto16 - 32 = 8a \\ \leadsto- 16 = 8a \\ \leadsto a = \frac{ - 16}{8} \\ \\ \fbox\bold{a = - 2 \: ms {}^{ - 2} } \\ \\ \fbox\bold{Force = mass \: \times acc {}^{n} } \\ \\ \bold{Force = 2 \times ( - 2)} \\ \\ \fbox \bold{Force = - 4 \: N}</p><p>$