Physics, asked by ps3110090, 7 months ago

A force of 100 N applied on an object accelerates it with an acceleration of 2ms-2. How much force is required to produce the same acceleration when an identical object of same mass are tied with the first object?


1.50 N
2.100 N
3.300 N
4.200 N

Answers

Answered by AnubhavRaj789guniess
21

Answer:

100 N

Explanation:

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Answered by BrainlyRonaldo
103

\bigstar Given

A force of 100 N applied on an object accelerates it with an acceleration of 2ms⁻²

\bigstar To Find

How much force is required to produce the same acceleration when an identical object of same mass are tied with the first object

\bigstar Solution

We know that

Newton's Second Law

\rm \longrightarrow F=ma

Here

  • F = force
  • m = mass
  • a = acceleration

Units

  • F = Newton (N)
  • m = kilogram (kg)
  • a = metres per second square (ms⁻²)

\star Finding mass

Therefore

\rm \longrightarrow m=\dfrac{F}{a}

Given that

A force of 100 N applied on an object accelerates it with an acceleration of 2ms⁻²

Hence

  • F = 100 N
  • a = 2 ms⁻²

Substituting the values

We get

\rm \longrightarrow m=\dfrac{100}{2} \ kg

Hence

\rm \longrightarrow m=50 \ kg

\star Finding Force

According to the question

We are asked to find the force is required to produce the same acceleration when an identical object of same mass are tied with the first object

Therefore

We must find force

Hence

Let the new force be F'

Therefore

\rm \longrightarrow F'=m'a'

Given that

A force of 100 N applied on an object accelerates it with an acceleration of 2ms⁻². How much force is required to produce the same acceleration when an identical object of same mass are tied with the first object

Here

...an identical object of same mass are tied with the first object

Hence

  • m' = 50 kg + 50 kg = 100 kg

Here

...same acceleration

Hence

Acceleration remains constant

Therefore

  • a' = a = 2 ms⁻²

Substituting the values

We get

\rm \longrightarrow F'=100 \times 2 \ N

Hence

\rm \longrightarrow F'=200 \ N

Therefore

\checkmark Option (4) is correct

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