Question

a contact force acts on an object of mass 5kg for a duration 2minutes .it increases the objects velocity from 18 m/s to 30 m/s find the magnitude.​

Answer 1

Answer :

F = 0.5 N

Explanation :

Given,

• mass of the body, m = 5 kg
• initial velocity, u = 18 m/s
• final velocity, v = 30 m/s
• time, t = 2 min

To find,

• magnitude of the force

Solution,

t = 2 min

t = 2 × 60 sec

t = 120 sec

     

we know,

• F = ma

where

 m - mass of the object

 a - acceleration of the object

acceleration is the rate of change of velocity

• a = (v - u)/t

   

Substituting the given values,

➙ a = (30 - 18)/120

➙ a = 12/120

➙ a = 1/10

➙ a = 0.1 m/s²

Substituting the values of m and a in F = ma,

➙ F = 5 kg × 0.1 m/s²

➙ F = 0.5 kg m/s²

➙ F = 0.5 N

∴ The magnitude of force is 0.5 N

Answer 2

Given :

• A contact force acts on an object of mass 5kg for a duration 2minutes .
• it increases the objects velocity from 18 m/s to 30 m/s.

To Find :

• The magnitude of the force applied = ?

Solution :

• Initial velocity (u) = 18 m/s
• Final velocity (v) = 30 m/s
• mass (m) = 5 kg
• Time (t) = 2 minutes

First of all convert the time taken from minutes to seconds :

→ Time = 2 minutes

→ Time = 2 × 60 seconds

→ Time = 120 seconds

• Hence,a contact force acts on an object of mass 5kg for a duration 120 seconds.

Finding the acceleration of the object :

→ Acceleration = (Final velocity - Initial velocity) ÷ Time

→ Acceleration = (30 - 18) ÷ 120

→ Acceleration = 12 ÷ 120

→ Acceleration = 0.1 m/s²

• Hence,the acceleration of the object is 0.1 m/s².

Now, let's find the magnitude of the force applied :

→ Force = mass × acceleration

→ Force = 5 × 0.1

→ Force = 0.5 N

• Hence,the magnitude of the force applied is 0.5 N.