constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s¯¹ to 3.5 m s-¯¹ in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force ?
Given:The mass of the body is 3.0 kg , theinitial speedof the body is 2.0 m / s , the final speedof the body is 3.5 m / s and the time taken is 25 s .
Acceleration produced in the body can be calculated usingthe first equation of motion,
v = u + a t
By substituting the given values in the above expression, we get
3.5 m / s = 2.0 m / s + a ( 25 s ) a = 1.5 m / s 25 s = 0.06 m / s
From Newton’s second law of the motion,
F = m a
By substituting the values in the above expression, we get
F = ( 3.0 kg ) ( 0.06 m / s ) = 0.18 N
Thus, aforce of 0.18 N is acting in the direction of motion of the body.
Given:The mass of the body is
Acceleration produced in the body can be calculated usingthe first equation of motion,
By substituting the given values in the above expression, we get
From Newton’s second law of the motion,
By substituting the values in the above expression, we get
Thus, aforce of
