a 10 kg mass moves along x-axis its acceleration as a function of its position is shown in the figure what is the total work done on the mass by the force as mass moves from x =0 to x=8 cm
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Here from the graph
a = mx
Where m is slope of the line
a = 20/8x = 2.5x
Now w = f *dx = ma*dx = 10*(2.5x)*dx
Now integrated and take limits 0 to8
W = 10*2.5/2 (8*8-0*0)=12.5*64=1000j
a = mx
Where m is slope of the line
a = 20/8x = 2.5x
Now w = f *dx = ma*dx = 10*(2.5x)*dx
Now integrated and take limits 0 to8
W = 10*2.5/2 (8*8-0*0)=12.5*64=1000j
devil1812:
8×10^-12 Joule is the answer
than a = 5/2x
now w = ma. dx = 10*5/2*x*dx
i have solved
25/2(0.08*0.08)=
got it
i have done now no need
max 20 already gave correct solution
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