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प्रश्न
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms–1. How long does the body take to stop?
उत्तर १
Retarding force, F = –50 N
Mass of the body, m = 20 kg
Initial velocity of the body, u = 15 m/s
Final velocity of the body, v = 0
Using Newton’s second law of motion, the acceleration (a) produced in the body can be computed as follows:
F = ma
–50 = 20 × a
`:. a = (-50)/20 = -2.5 "m/s"^2`
The first equation of motion allows for the calculation of the body's time (t) to come to rest as follows:
v = u + at
`:. t = (-u)/a = (-15)/(-2.5) = 6 s`
उत्तर २
Here m = 20 kg, F = – 50 N (retardation force)
As F = ma
`=> a = F/m = (-50)/20 = -2.5 ms^(-2)`
Using equation v = u + at
Given `u = 15 ms^(-1), v = 0`
Now, 0 = 15 + (-25) t
or t = 6 s
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