Question

a constant force acts on an object of mass 5kg for a duration for 2s. it increases the object velocity from 3m/s to 7m/s. find the magnitude of the force applied force. now if the force was applied for a duration of 5s, what would be the final velocity of the object?

Answer 1

Answer:

i) 10N

ii) 17m/s

Explanation:

hopefully this may help

Answer 2

$\huge\mathfrak\green {Solution}$

★ Given :-

${ \ U \ = \ 3 m s¯¹}$

${ \ V \ = \ 7m s¯¹}$

${ \ T \ = \ 2s}$

${ \ M \ = \ 5kg}$

★ We have :-

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎${ \ F \ =}$ $\frac{m(v-u)}{t}$

Substitution of values in this relation gives

${ \ F \ = \ 5kg}$ $\frac{(7m s¯¹ - 3m s¯¹)}{2s = 10 N}$

★ Now,

If this force is applied for a duration of 5s (t = 5s), then the final velocity can be calculated by rewriting

‎

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎${ \ v \ = \ u \ +}$ $\frac{Ft}{m}$

On substituting the values of u, F, m and t, we get the final velocity.

‎ ‎ ‎ ‎${ \ v \ = \ 13m s¯¹}$