A 3.0N force to the right acts on a 0.5kg object at rest during a time interval of 4.0 seconds .What is the velocity of the object at the end of this interval?

24 m/s

Explanation:

Applying Newton's second law of motion,

F = ma Equation 1

Where F = force acting on the object, m = mass of the object a = acceleration of the object.

First we calculate for a.

make a the subject of the formula in equation 1 above

a = F/m Equation 2

Given: F = 3 N, m = 0.5 kg.

Substitute into equation 2

a = 3/0.5

a = 6 m/s²

Secondly we calculate the velocity by using,

v = u+at Equation 3

Where v = Final velocity of the object, u = initial velocity of the object, t = time interval

Given: u = 0 m/s (at rest), a = 6 m/s², t = 4 s

Substitute into equation 3

v = 0+6(4)

v = 24 m/s

see earlier