Question

a force =10x^2 + 0.2x act on a particle in x direction , find the work done by it force during a displacement from x=0 to x=4
​

Answer 1

Answer:

Small amount of work done dW in giving a small displacement d

x

is given by

dW=

x

or dW=Fdxcos0

∘

or dW=Fdx[∴cos0

∘

=1]

Total work done, W = ∫

x=0

x=2

Fdx=∫

x=0

x=2

(0.5x+10)dx

=∫

x=0

x=2

0.5xdx+∫

x=0

x=2

10dx=0.5

∣

∣

∣

∣

∣

2

x

2

∣

∣

∣

∣

∣

x=0

x=2

+10∣x∣

x=0

x=2

=

2

0.5

[2

2

−0

2

]+10(2−0]=(1+20)=21J

solution

Explanation:

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