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An object of mass 10 kg, initially at rest, experiences a constant horizontal acceleration of 4 m/s2 due to the action of a resultant force applied for 20 s. Determine the total amount of energy transfer by work, in kj.

Sagot :

MDKPfromIDNidn

So, the total energy value that has been transferred by work is 32 kJ.

Introduction

Hi ! In this question, I will help you. Work is the amount of force exerted to cause an object to move a certain distance from its starting point. In physics, the amount of work will be proportional to the increase in force and increase in displacement. Amount of work can be calculated by this equation :

[tex] \boxed{\sf{\bold{W = F \times s}}} [/tex]

With the following condition :

  • W = work (J)
  • F = force (N)
  • s = shift or displacement (m)

Now, because in this question, the "s" is not directly known, whereas it is known that the initial velocity is zero, the object has an acceleration, and is moving in certain time intervals. Then, use this formula to find the value of "s" !

[tex] \boxed{\sf{\bold{s = \frac{1}{2} \times a \times t^2}}} [/tex]

With the following condition :

  • s = shift or displacement (m)
  • a = acceleration (m/s²)
  • t = interval of the time (s)

Problem Solving

We know that :

  • a = acceleration = 4 m/s²
  • t = interval of the time = 20 s
  • m = mass = 10 kg

What was asked :

  • W = work = ... kJ

Step by Step :

  • Calculate value of force by multiple m and a

[tex] \sf{F = m \times a} [/tex]

[tex] \sf{F = 10 \times 4} [/tex]

[tex] \sf{\bold{F = 40 \: N}} [/tex]

  • Calculate value of shift or displacement

[tex] \sf{s = \frac{1}{2} \times a \times t^2} [/tex]

[tex] \sf{s = \frac{1}{\cancel 2} \times \cancel 4 \:_2 \times 20^2} [/tex]

[tex] \sf{s = 2 \times 400} [/tex]

[tex] \sf{\bold{s = 800 \: m}} [/tex]

  • Last, calculate the value of work

[tex] \sf{W = F \times s} [/tex]

[tex] \sf{W = 40 \times 800} [/tex]

[tex] \boxed{\sf{W = 32,000 \: J = 32 \: kJ}} [/tex]

So, the total energy value that has been transferred by work is 32 kJ.

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