A 30 kg bicycle is at rest. A 60 N force to the right acts on the bicycle during a time interval of 1.5 seconds. What is the velocity of the bicycle at the end of this interval?



Answer :

According to Newton's second law,

[tex]F=ma[/tex]

Where F is the force acting on a body, m is the mass of the body, and a is the acceleration of the body due to the applied force.

The acceleration is given by,

[tex]a=\frac{v}{t}[/tex]

Where v is the velocity and t is the time interval.

Substituting this equation in the first equation,

[tex]F=\frac{mv}{t}[/tex]

From the question,

F=60 N

m=30 kg

t=1.5 s

On substituting these values in the equations,

[tex]60=30\times\frac{v}{1.5}[/tex]

Which gives,

[tex]v=\frac{60\times1.5}{30}=3\text{ m/s}[/tex]

Therefore the velocity of the bicycle is 3 m/s

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