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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?

Respuesta :

ujalakhan18

Answer:

[tex]\huge\boxed{\sf t = 0.9 \ s}[/tex]

Explanation:

Given Data:

Mass = m = 1.2 kg

Initial Velocity = [tex]V_i[/tex] = 2.0 m/s

Final Velocity = [tex]V_f[/tex] = 5.0 m/s

Force = F = 4.0 N

Required:

Time = t = ?

Formula:

[tex]\displaystyle F=\frac{m(V_f-V_i)}{t}[/tex]

Solution:

Put the given data in the above formula

[tex]\displaystyle 4=\frac{1.2(5-2)}{t} \\\\4=\frac{1.2(3)}{t} \\\\4=\frac{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}[/tex]

tanweeleong195oxyqwp

Answer:

0.9s

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 =

[tex]3 \frac{1}{3} \frac{m}{ {s}^{2} } [/tex]

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

v = u + at,

[tex]5 = 2 + (3 \frac{1}{3} )t \\3 \frac{1}{3} t = 3 \\ t = 3 \div 3 \frac{1}{3} = 0.9s[/tex]

Therefore time taken = 0.9s

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