Question

GRAPHICAL \& INTEGRAL METHODS 6 A force of 5.0 N was applied to a particle to move it 4 meters to the right. The applied force decreases to zero in the next 2 meters. Calculate the work done by the force on the particle. SOLUTION (GRAPHICAL) The net work done by this force \[ \begin{aligned} W_{\text {Ato } \mathrm{C}}= & W_{\mathrm{Ato} \mathrm{B}}+W_{\mathrm{B} \text { to } \mathrm{C}} \\ & W_{\text {A to B }}=(5.0 \mathrm{~N})(4.0 \mathrm{~m})=20 \mathrm{~J} \\ & W_{\mathrm{B} \text { to } \mathrm{C}}=\frac{1}{2}(5.0 \mathrm{~N})(2.0 \mathrm{~m})=5.0 \mathrm{~J} \\ = & 20 \mathrm{~J}+5.0 \mathrm{~J}=25 \mathrm{~J} \end{aligned} \]

          GRAPHICAL \& INTEGRAL METHODS
6
A force of 5.0 N was applied to a particle to move it 4 meters to the right. The applied force decreases to zero in the next 2 meters. Calculate the work done by the force on the particle.

SOLUTION (GRAPHICAL)

The net work done by this force
\[
\begin{aligned}
W_{\text {Ato } \mathrm{C}}= & W_{\mathrm{Ato} \mathrm{B}}+W_{\mathrm{B} \text { to } \mathrm{C}} \\
& W_{\text {A to B }}=(5.0 \mathrm{~N})(4.0 \mathrm{~m})=20 \mathrm{~J} \\
& W_{\mathrm{B} \text { to } \mathrm{C}}=\frac{1}{2}(5.0 \mathrm{~N})(2.0 \mathrm{~m})=5.0 \mathrm{~J} \\
= & 20 \mathrm{~J}+5.0 \mathrm{~J}=25 \mathrm{~J}
\end{aligned}
\]

GRAPHICAL & INTEGRAL METHODS
6
A force of 5.0 N was applied to a particle to move it 4 meters to the right. The applied force decreases to zero in the next 2 meters. Calculate the work done by the force on the particle.

SOLUTION (GRAPHICAL)

The net work done by this force

WAto C= WAtoB+WB to C
WA to B =(5.0 N)(4.0 m)=20 J
WB to C=(1)/(2)(5.0 N)(2.0 m)=5.0 J

= 20 J+5.0 J=25 J

Added by Santiago A.

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