Question

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Answer 1

Using the first equation of motion

v = u + at

See above solution

Hope it helps you ❣️☑️☑️

Answer 2

Q. A constant force acts on an object of mass 2kg for 10 sec. And increase its velocity for 5m/s to 10m/s. Find the magnitude of the applied force If the force was for a duration of 15 sec. what would be the velocity of the object.

${\bold {\huge {\pink {Answer :}}}}$

Given:

mass = 2 kg

time(t) =10 s

initial velocity(u) = 5 m/s

final velocity(v) = 10 m/s

so let the acceleration be a

be

$a = \frac{v - u}{t} = \frac{10 - 5}{10} \\ = 0.5 \: \: m / {s}^{2}$

now,

Magnitude of force applied will be

F= ma

$F= ma \: \\ F= 2 \times 0.5 \\ F= 1.0 \: N$

and the final velocity after 15 s is v

so

v = u + at

v = 5 + 0.5 x 15

v = 12.5 m/s

So the velocity of the object would be 12.5 m/s.