A vehicle of mass 20 kg is moving with a velocity of 4 m/s.Find the magnitude of the force that is to be applied on the vehicle so that the vehicle have a velocity of 1 m/s after travelling a distance of 20 m.
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Retardation of vehicle is
a = (u^2 - v^2) / (2 S)
= (4^2 - 1^2) / (2 × 20)
= 0.375 m/s^2
F = ma
= 20 kg × 0.375 m/s^2
= 7.5 N
Magnitude of Force to be applied is 7.5 N
a = (u^2 - v^2) / (2 S)
= (4^2 - 1^2) / (2 × 20)
= 0.375 m/s^2
F = ma
= 20 kg × 0.375 m/s^2
= 7.5 N
Magnitude of Force to be applied is 7.5 N
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hello,
given mass is 20kg initial velocity is 4m/s and final velocity is 1m/s
distance travelled is 20m
by using equations of motion,we will use 3rd equation
2aS=v²-u² where a is acceleration S is distance travelled v is final velocity u is initial velocity
2a20=1²-4²
40a=1-16
a=-15/40=-0.375 m/s² is acceleration -ve sigh shows that it is slowing down of deceleration
now, F=ma where F is force applied m is mass of the body and a is acceleration
∴F=20×0.375=7.5N
force applied to change velocity from 4m/s to 1m/s is 7.5N
given mass is 20kg initial velocity is 4m/s and final velocity is 1m/s
distance travelled is 20m
by using equations of motion,we will use 3rd equation
2aS=v²-u² where a is acceleration S is distance travelled v is final velocity u is initial velocity
2a20=1²-4²
40a=1-16
a=-15/40=-0.375 m/s² is acceleration -ve sigh shows that it is slowing down of deceleration
now, F=ma where F is force applied m is mass of the body and a is acceleration
∴F=20×0.375=7.5N
force applied to change velocity from 4m/s to 1m/s is 7.5N
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