Question

4. A constant force acts on an object of mass 2 kg for a duration of 5 seconds. It increases the object's velocity from 3m/s to 8m/s. Find the applied force.​

Answer 1

GIVEN :-

• Mass of the object ( m ) = 2 kg.
• Duration of time ( t ) = 5 seconds.
• Initial velocity ( u ) = 3 m/s.
• Final velocity ( v ) = 8 m/s.

TO FIND :-

• The Applier Force ( F ).

CALCULATION :-

This question can be solved with 2 different methods, let's begin with first one,

Using Newton's second law of motion,

→ F = ma

→ F = m × ( v - u )/t. [ a = ( v - u )/t ]

→ F = 2 kg × ( 8 - 3 )/5

→ F = 2kg × 5/5 m/s²

→ F = 2kg × 1 m/s²

→ F = 2 kg.m/s²

→ F = 2 Newtons.

Now let's start second method ,

As we know that , the applied force is equal to the rate of change of momentum .

→ F = ∆p/∆t

→ F = ( mv - mu )/t

→ F = [ m( v - u )/t ]

→ F = 2 ( 8 - 3 )/t

→ F = 10/5 kg.m/s²

→ F = 2 Newtons.

.°. The applied force is 2 Newtons.

Answer 2

Explanation:

given :

• mass = 2 kg

• time = 5 sec

• first velocity = 3 m/s

• last velocity = 8 m/s

to find :

• Find the applied force.

knowledge required :

• velocity of the objects is given, but we have to assume that initial velocity and final velocity. From the mass and acceleration of the object, we have to find the force of the object.

• The expression of finding acceleration is

• a= ( v-u)/t

• Where, a be the acceleration, u be the initial

• velocity, u be the final velocity and t be the The expression of finding final velocity is v = u + at. According to the newton second law of motion the expression of finding force is

• time

• f = ma

• Here, m be the mass and f be the force.

solution :

• first solution we will do

• with newton method

• force = ma

• force = 2 x (8-3)/5

• force = 2x 5/5 m/s²

• force = 2 x 1 m/s²

• force = 2 m/s²

• in first method the answer is = 2 m/s

second method solution we will do:

• force = Ap/At

• Force = (mv - mu )/t

• Force=[ m(v - u)/t]

• Force = 2 (8-3)/t

• Force = 10/5 m/s²

• then we will divide 10÷5

• the answer is = 2

• so the object applied velocity = 2 m/s