Question

please do it as fast as possible...​

Answer 1

Question: A force of 15 Newton is applied to a body of mass 3 kg initially at rest on a smooth surface for 3 seconds. Calculate the final velocity, the distance travelled and the work done.

• So sorry dear user but I don't know what is the last part of this question to solve as it isn't clearly visible.

• Edit: May be it is final kinetic energy, so if I m thinking right then you can see below for this part too! (:

Provided that:

• Force = 15 N
• Initial velocity = 0 m/s
• Mass = 3 kg
• Time = 3 seconds

To calculate:

• The final velocity
• The distance travelled
• The work done
• Final kinetic energy

Solution:

• The final velocity = 15 m/s
• The distance = 22.5 m
• The work done = 337.5 J
• Final kinetic energy = 337.5 J

Using concepts:

• First equation of motion
• Force formula
• Third equation of motion
• Work formula
• Kinetic energy formula

Using formulas:

• First equation of motion,

• ${\small{\underline{\boxed{\pmb{\sf{v \: = u \: + at}}}}}}$

• Third equation of motion,

• ${\small{\underline{\boxed{\pmb{\sf{v^2 \: - u^2 = 2as}}}}}}$

• Second law of motion that tell us about force is given by,

• ${\small{\underline{\boxed{\pmb{\sf{F \: = ma}}}}}}$

• Work is given by,

• ${\small{\underline{\boxed{\pmb{\sf{W \: = Fs}}}}}}$

• Kinetic energy formula:

• ${\small{\underline{\boxed{\pmb{\sf{E_k \: = \dfrac{1}{2} \: mv^2}}}}}}$

Where, a denotes acceleration, u denotes initial velocity, s denotes distance, Displacement or height, W denotes work, F denotes force, m denotes mass v denotes final velocity, E_ k denotes kinetic energy and t denotes time.

Required solution:

~ Firstly to find out the final velocity we need the acceleration so let us calculate the acceleration first by using second law of motion, force formula!

$:\implies \sf Force \: = Mass \times Acceleration \\ \\ :\implies \sf F \: = m \times a \\ \\ :\implies \sf F \: = ma \\ \\ :\implies \sf 15 = 3 \times a \\ \\ :\implies \sf \dfrac{15}{3} \: = a \\ \\ :\implies \sf 5 = \: a \\ \\ :\implies \sf a \: = 5 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 5 \: ms^{-2}$

~ Now let's find out the final velocity by using first equation of motion!

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf v \: = 0 + 5(3) \\ \\ :\implies \sf v \: = 0 + 15 \\ \\ :\implies \sf v \: = 15 \: ms^{-1} \\ \\ :\implies \sf Final \: velocity \: = 15 \: ms^{-1}$

~ Now let's calculate the distance by using third equation of motion!

$:\implies \sf v^2 \: - u^2 = 2as \\ \\ :\implies \sf (15)^{2} - (0)^{2} = 2(5)(s) \\ \\ :\implies \sf 225 - 0 = 10s \\ \\ :\implies \sf 225 = 10s \\ \\ :\implies \sf \dfrac{225}{10} \: = s \\ \\ :\implies \sf 22.5 \: = s \\ \\ :\implies \sf Distance \: = 22.5 \: metres$

~ Now let's find out the work done!

$:\implies \sf W \: = Fs \\ \\ :\implies \sf W \: = 15(22.5) \\ \\ :\implies \sf W \: = 15 \cdot 22.5 \\ \\ :\implies \sf W \: = 337.5 \: Joules \\ \\ :\implies \sf Work \: done \: = 337.5 \: Joules$

~ Finding the final kinetic energy!

$:\implies \sf E_k \: = \dfrac{1}{2} \: mv^2 \\ \\ :\implies \sf E_k \: = \dfrac{1}{2} \times 3 \times (15)^{2} \\ \\ :\implies \sf E_k \: = \dfrac{1}{2} \times 3 \times 225 \\ \\ :\implies \sf E_k \: = \dfrac{1}{2} \times 675 \\ \\ :\implies \sf E_k \: = 337.5 \: J \\ \\ :\implies \sf Kinetic \: energy \: = 337.5 \: Joules$

Hope it's helpful! :)