Question

2. A constant force acts on an object of mass 10 kg for a duration of 2 s. It increases the
object's velocity from 7 ms-1 to 11 m 5-1. The Force applied is equal to newtons.​

Answer 1

Answer:

Given,

mass=5kg

t

1

=2s

Initial velocity u=3m/s

Final velocity v=7m/s

t

2

=5s

So,

Let the Force be F

Let the acceleration be a

So,

a=

t

(v−u)

=

2

(7−3)

=2m/s

2

So the magnitude of the applied force is 10N

And the final velocity after 5s is v

So,

v=u+at

v=3+2×5

v=13m/s

The final velocity after 5s is 13m/s

Answer 2

Given:-

• Mass (m) = 10kg

• Time taken (t) = 2s

• Initial velocity (u) = 7m/s

• Final velocity (v) = 11m/s

To Find:-

• Force applied (F)

Solution:-

Firstly we calculate the acceleration of the object

So by using 1st equation of motion

→ v = u+at

Substitute the value we get

→ 11 = 7 + a×2

→ 11-7 = a×2

→ 4 = a×2

→ a = 4/2

→ a = 2m/s

Now Calculating the Force

→ F = m×a

Substitute the value we get

→ F = 10×2

→ F = 20N

∴ The force applied on it is 20Newton.