A 10kg 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 the mass moves from $x = 0$ to $x = 8\,cm$
A 10kg 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 the mass moves from $x = 0$ to $x = 8\,cm$
$8 \times {10^{ - 2}}joules$
$16 \times {10^{ - 2}}joules$
$4 \times {10^{ - 4}}joules$
$1.6 \times {10^{ - 3}}joules$
A 10kg 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 the mass moves from $x = 0$ to $x = 8\,cm$
Work done = Area covered in between force displacement curve and displacement axis
= Mass $\times$ Area covered in between acceleration-displacement curve and displacement axis.
= $10 \times \frac{1}{2}(8 \times {10^{ - 2}} \times 20 \times {10^{ - 2}})$
= $8 \times {10^{ - 2}}\,J$
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