Question

3. The velocity of a particle of mass 150 g changes from 8 m/s to 12 m/s in two seconds. Assuming that a constant force acts on it, find the magnitude of the force.​

Answer 1

Answer:

Provided that:

• Initial velocity = 8 m/s
• Time = 2 seconds
• Final velocity = 12 m/s
• Mass = 150 g

To calculate:

• The force

Solution:

• The force = 0.30 Newton

Knowledge required:

• SI unit of mass = kg
• SI unit of force = Newton (N)
• SI unit of time = second (s)
• SI unit of velocity (m/s)

Using concepts:

• Formula to convert g into kg
• Force formula

Using formulas:

• ${\small{\underline{\boxed{\pmb{\sf{F \: = m \cdot \bigg \lgroup \dfrac{v-u}{t} \bigg \rgroup}}}}}}$

Where, a denotes acceleration, u denotes initial velocity, F denotes force v denotes final velocity, m denotes mass and t denotes time taken.

• ${\small{\underline{\boxed{\pmb{\sf{1 \: g \: = \dfrac{1}{1000} \: kg}}}}}}$

Required solution:

~ Firstly let us convert grams into kilograms by using suitable formula!

$:\implies \sf 1 \: g \: = \dfrac{1}{1000} \: kg \\ \\ :\implies \sf 150 \: g \: = \dfrac{150}{1000} \: kg \\ \\ :\implies \sf 150 \: g \: = \dfrac{15}{100} \: kg \\ \\ :\implies \sf 150 \: g \: = 0.15 \: kg \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

~ Now let's calculate force!

$:\implies \sf F \: = m \cdot \bigg \lgroup \dfrac{v-u}{t} \bigg \rgroup \\ \\ :\implies \sf F \: = 0.15 \cdot \bigg \lgroup \dfrac{12-8}{2} \bigg \rgroup \\ \\ :\implies \sf F \: = 0.15 \cdot \bigg \lgroup \dfrac{4}{2} \bigg \rgroup \\ \\ :\implies \sf F \: = 0.15 \cdot 2 \\ \\ :\implies \sf F \: = 0.30 \: Newton \\ \\ :\implies \sf Force \: = 0.30 \: Newton$