A force acts on a $2\ kg$ object so that its position is given as a function of time as $x = 3t^2 + 5$. What is the work done by this force in first $5$ seconds ?

Q: A force acts on a \$2\ kg\$ object so that its position is given as a function of time as \$x = 3t^2 + 5\$. What is the work done by this force in first \$5\$ seconds ?

\$x = 3t^2 + 5\$\$v = \dfrac{dx}{dt}\$\$v = 6t + 0\$\$at \, t = 0 \$ \$v = 0\$t = 5 sec v = 30 m/sW.D. = \$\Delta KE\$\$W.D. = \dfrac{1}{2} mv^2 - 0 = \dfrac{1}{2} (2) (30)^2 = 900 J\$

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Accepted Answer

$x = 3t^2 + 5$$v = \dfrac{dx}{dt}$$v = 6t + 0$$at \, t = 0 $   $v = 0$t = 5 sec   v = 30 m/sW.D. = $\Delta KE$$W.D. = \dfrac{1}{2} mv^2 - 0 = \dfrac{1}{2} (2) (30)^2 = 900 J$

Work Energy And Power

Physics
4882 Views

A force acts on a $2 k g$ object so that its position is given as a function of time as $x = 3 t^{2} + 5$ . What is the work done by this force in first $5$ seconds ?

Work Energy And Power

Physics 4882 Views

A force acts on a 2 kg object so that its position is given as a function of time as x=3t2+5. What is the work done by this force in first 5 seconds ?

Physics
4882 Views

A force acts on a $2 k g$ object so that its position is given as a function of time as $x = 3 t^{2} + 5$ . What is the work done by this force in first $5$ seconds ?