Question

A force of 20N acts on a body of mass 1 kg at rest. The
work done in 8 sec. is
(1) 6400 J
(2) 3200 J
(3) 12800 J
(4) 19200 J​

Answer 1

Given :

• Force = 20 N
• Mass = 1 kg
• Initial velocity = 0 m/s

To find :

• Work done in 8 sec

Solution :

$\large \boxed{ \sf \red{ F = ma}} \\ \\ \sf a = \frac{f}{m} \\ \\ \sf a = \frac{20}{1} \\ \\ \sf a = 20 \: m {s}^{ - 2}$

$\large\boxed{ \sf \blue{ s = ut + \frac{1}{2} a {t}^{2}} } \\ \\ \sf s = 0 \times 8 + \frac{1}{2} \times 20 \times {8}^{2} \\ \\ \sf s = 640 \: m$

$\boxed{\sf \orange {Work = Force Displacement}} \\ \\ \sf Work \: = 20 \times 640 \\ \\ \large \boxed{\sf Work = 12800 \: J}$

Answer 2

$\large{\underline{\underline{\mathfrak{Answer :}}}}$

(c) 12,800 J

$\rule{200}{0.5}$

$\underline{\underline{\mathfrak{Step-By-Step-Explanation :}}}$

Given :

• Force is 20 N
• Initial Velocity (u) = 0 m/s
• Mass of Object = 1 kg
• Time (t) = 8 s

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To Find :

• Work Done

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Solution :

We have formula for Force :

$\large{\boxed{\sf{F \: = \: ma}}} \\ \\ \rightarrow {\sf{a \: = \: \dfrac{F}{m}}} \\ \\ \rightarrow {\sf{a \: = \: \dfrac{20}{1}}} \\ \\ \underline{\sf{\therefore \: Acceleration \: is \: 20 \: ms^{-2}}}$

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And Now, use 2nd equation of motion :

$\large{\boxed{\sf{S \: = \: ut \: + \: \dfrac{1}{2} at^2}}} \\ \\ \rightarrow {\sf{S \: = \: 0 \: \times \: 3 \: + \: \dfrac{1}{2} \: \times \: 20 \: \times \: 8^2}} \\ \\ \rightarrow {\sf{S \: = \: 10 \: \times \: 64}} \\ \\ \rightarrow {\sf{S \: = \: 640}} \\ \\ \underline{\sf{\therefore \: Displacement \: is \: 640 \: m}}$

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We have formula for Work Done :

$\large{\boxed{\sf{W \: = \: F.s}}} \\ \\ \rightarrow {\sf{W \: = \: 20 \: \times \: 640}} \\ \\ \rightarrow {\sf{W \: = \: 12800}} \\ \\ \underline{\sf{\therefore \: Work \: Done \: is \: 128000 \: J}}$