Question

An object of mass 1.2 kg is moving with a velocity of 2.0m/s when it is acted on by a force of

4.0 N. The velocity of the object increases to 5.0 m/ S.
For what period of time does the force act on the obiect?

Answer 1

0.9s

Step-by-step explanation:

Solution

This question tests on the concept of the Kinematics Equations.

Given information from the question,

we can use the equation:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.

We also need the equation: F = ma,

where F is the force applied, m is the mass the object, and a is acceleration.

We first find the acceleration of the object.

F = ma,

4N = (1.2kg) × a

a = 4 ÷ 1.2 =

`3 (1)/(3) \frac{m}{ {s}^(2) }`

Now with the acceleration, we can find the time taken for the acceleration to take effect.

v = u + at,

`5 = 2 + (3 (1)/(3) )t \\3 (1)/(3) t = 3 \\ t = 3 / 3 (1)/(3) = 0.9s`

Therefore time taken = 0.9s

Answer 2

`\huge\boxed{\sf t = 0.9 \ s}`

Step-by-step explanation:

Given Data:

Mass = m = 1.2 kg

Initial Velocity =
`V_i` = 2.0 m/s

Final Velocity =
`V_f` = 5.0 m/s

Force = F = 4.0 N

Required:

Time = t = ?

Formula:

`\displaystyle F=(m(V_f-V_i))/(t)`

Solution:

Put the given data in the above formula

`\displaystyle 4=(1.2(5-2))/(t) \\\\4=(1.2(3))/(t) \\\\4=(3.6)/(t) \\\\Multiply \ t\ to \ both \ sides\\\\4t = 3.6\\\\Divide \ both \ sides \ by \ 4\\\\t=3.6/4\\\\t = 0.9 \ s \\\\\rule[225]{225}{2}`