Question

a constant force acts on an object of mass 5 kg for a duration of 2 s. it increases the object`s velocity from 3 m s.1 to 7 m s-1. find the magnitude of the applied force . now, if the force was applied for duration of 5s, what would be final velocity of the object?

Answer 1

Since the force on the object is constant, its acceleration is also constant.

Acceleration is found using,

$v=u+at\\ 7=3+2a\\ a=2$

The magnitude of the applied force is

$F=ma\\ F=5(2)\\ F=10 \;N$

Velocity of the object after 5 seconds is

$v=2+(2)5\\ v=12 \;m/s$

Answer 2

The magnitude of the applied force is 10 N and final velocity 13 m/s.

Explanation:

Given mass of the object, m = 5 kg.

Initial velocity of the object, u = 3 m/s

Final velocity of the object, v = 7 m/s.

Time taken to reach the final velocity, t = 2 s.

Use first equation of motion,

v = u + at

substitute the values, we get

7 m/s = 3 m/s + a x 2 s

$a = 2m/s^2$

So, constant force

F =ma = 2 x 5 = 10 N.

As force is constant, acceleration is also constant.

If the force was applied for duration of 5s, then final velocity becomes

V = 3 m/s + 2 x 5

V = 13 m/s.

Thus, the magnitude of the applied force is 10 N and final velocity 13 m/s.

# Learn More: equation of motion.

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