Question

A force of (2i+3j)N acts on a body of mass 1 kg which is at rest initially. What is the acceleration of the body?​

Answer 1

A force of (2i + 3j) N acts on a body of mass 1kg which is at rest initially.

To find : The acceleration of the body.

solution : according to Newton's law of motion, force is the product of mass and acceleration.

i.e., F = ma , where m is mass of body and a is acceleration of that body.

here F = (2i + 3j) N , m = 1 kg

so, acceleration of the body, a = F/m

= (2i + 3j)/1

= (2i + 3j) m/s²

Therefore the magnitude of acceleration is √(2² + 3²) = √(13) = 3.6 m/s².

also read similar questions : a body of mass 5 kg is pulled by a force F=(-4i+3j) N from rest on a smooth horizontal table in x z plane the time at wh...

https://brainly.in/question/9016474

what is the torque of the force f=(2i-3j+4k)F, acting at the point r=(3i+2j+3k)m ABOUT THE ORIGIN (in N-m)

https://brainly.in/question/4290254

Answer 2

$\large{\bf{\red{\underline{\underline{Answer:}}}}}$

Given,

• Force = (2i+3j)N
• Mass = 1kg

To Find:-

• Acceleration of a body.

Formula used:-

F = ma

F = Force

m=mass

a= acceleration.

Calculations:-

F =ma

a=F/m

now substitute the given values:-

= (2i + 3j)/1

= (2i + 3j) m/s²

=√(2² + 3²)

=.3.6 m/s²

The acceleration of the body is 3.6 m/s².