Question

A constant force acting on a body of mass 3kg change its speed from 2m/s to 3.5m/s in 25 sec. In the direction of motion of the body what is the magnitude and direction of force

Answer 1

v=3.5 m/s
u=2 m/s
t=25 sec
m=3 kg
F=ma or F=m x a
a=v-u÷t
F=3 x (*3.5 - 2 ÷ 25)< *THE WHOLE THING IN THE BRACKET IS DIVIDED BY 25 >
F=3 x 1.5 ÷ 25
F=4.5 ÷ 25
F=0.18N
If you consider the object moving in the east direction (initially) then it is east.(as it is positive)
If you consider the object moving in the west direction (initially) then it is west.(as it is positive)

Answer 2

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ANSWER:-

0.18 N; in the direction of motion of the body

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EXPLANATION:--

Mass of the body, m = 3 kg

Initial speed of the body, u = 2 m/s

Final speed of the body, v = 3.5 m/s

Time, t = 25 s

Using the first equation of motion, the acceleration (a) produced in the body can be calculated as:

v = u + at

=> a =( v - u)/t

= (3.5 - 2)/ 25

= 1.5/ 25

= 0.06 m/s^2

As per Newton’s second law of motion, force is given as:

F = ma = 3 × 0.06 = 0.18 N

Since the application of force does not change the direction of the body, the net force acting on the body is in the direction of its motion.

I hope, this will help you

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