Question

A body of mass 5kg, initially at rest, is moved by a horizontal force of 2N on
smooth horizontal surface. Find the work done by the force in 10s?​

Answer 1

Answer:

the answer is 17n

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Answer 2

Answer:

work done = 40 J

Explanation:

$W = F S\\\\F = M A\\\\a= \frac{F}{M} \\\\a = 2/5 = 0.4 m/s^{2}$

$v = u + at\\ v = 0 + 0.4 * 10s\\\\v = 4 m/s$

$v^{2} - u^{2} = 2 a s\\\\s= 16/2(0.4) = 20$

$W = F S \\W = 2 N * 20 m = 40 J$