Question

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?

Answer 1

Answer:

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

Answer 2

Answer:

v = 24m/s

Explanation:

To find the velocity of the object you can use the Newton second law, and the formula for the acceleration in an accelerated motion:

$F=ma\\\\a=\frac{v_f-v_o}{t}$

F: force = 3.0N

a: acceleration

m: mass of the object = 0.5kg

vf: final velocity

vo: initial velocity = 0 m/s

t: time in which the force is applied = 4.0s

From the first equation you can calculate the acceleration of the object. With the value of a you can calculate the final velocity. Hence, by replacing the values of F, m, vo and t you obtain:

$a=\frac{F}{m}=\frac{3.0N}{0.5kg}=6\frac{m}{s^2}\\\\v_f=at+v_o=(6\frac{m}{s^2})(4.0s)+0\frac{m}{s}=24\frac{m}{s}$

hence, the velocity of the object is 24m/s