please do it as fast as possible...
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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.
Provided that:
To calculate:
Solution:
Using concepts:
Using formulas:
• First equation of motion,
• Third equation of motion,
• Second law of motion that tell us about force is given by,
• Work is given by,
• Kinetic energy formula:
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!
[tex]:\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}[/tex]
~ Now let's find out the final velocity by using first equation of motion!
[tex]:\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}[/tex]
~ Now let's calculate the distance by using third equation of motion!
[tex]:\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[/tex]
~ Now let's find out the work done!
[tex]:\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[/tex]
~ Finding the final kinetic energy!
[tex]:\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[/tex]
Hope it's helpful! :)