355834
A 10 \(kg\) brick moves along an \(x\)-axis. Its acceleration as a function of its position is shown in figure. What is the net work performed on the brick by the force causing the acceleration as the brick moves from \(x=0\) to \(x = 8.0\,m\) ?
1 4\(J\)
2 8\(J\)
3 2\(J\)
4 1\(J\)
Explanation:
According to the graph the acceleration \(a\) varies linearly with the coordinate \(x\). We may write \(a=\alpha x\), where \(\alpha\) is the slope of the graph \(\alpha = \frac{{20}}{8} = 2.5\;m{s^{ - 1}}\) The force on the brick is in the positive \(x\) direction and according to Newton's second law, its magnitude is given by \(F=\dfrac{a}{m}=\dfrac{\alpha}{m} x\) if \(x_{f}\) is the final coordinate, the work done by the force is \(\begin{aligned}W & =\int_{0}^{x_{f}} F d x=\dfrac{\alpha}{m} \int_{0}^{x_{f}} x d x \\& =\dfrac{\alpha}{2 m} x_{f}^{2}=\dfrac{2.5}{2 \times 10} \times(8)^{2}=8 J\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355835
A particle of mass 6 \(kg\) moves according to the law \(x=0.2 t^{2}+0.02 t^{3}\). Find the work done by the force in first 4 \(s\).
355836
When a rubber-band is stretched by a distance \(x\), it exerts a restoring force of magnitude \(F=a x+b x^{2}\) where \(a\) and \(b\) are constants. The work done in stretching the unstrected rubber band by \(L\) is
355834
A 10 \(kg\) brick moves along an \(x\)-axis. Its acceleration as a function of its position is shown in figure. What is the net work performed on the brick by the force causing the acceleration as the brick moves from \(x=0\) to \(x = 8.0\,m\) ?
1 4\(J\)
2 8\(J\)
3 2\(J\)
4 1\(J\)
Explanation:
According to the graph the acceleration \(a\) varies linearly with the coordinate \(x\). We may write \(a=\alpha x\), where \(\alpha\) is the slope of the graph \(\alpha = \frac{{20}}{8} = 2.5\;m{s^{ - 1}}\) The force on the brick is in the positive \(x\) direction and according to Newton's second law, its magnitude is given by \(F=\dfrac{a}{m}=\dfrac{\alpha}{m} x\) if \(x_{f}\) is the final coordinate, the work done by the force is \(\begin{aligned}W & =\int_{0}^{x_{f}} F d x=\dfrac{\alpha}{m} \int_{0}^{x_{f}} x d x \\& =\dfrac{\alpha}{2 m} x_{f}^{2}=\dfrac{2.5}{2 \times 10} \times(8)^{2}=8 J\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355835
A particle of mass 6 \(kg\) moves according to the law \(x=0.2 t^{2}+0.02 t^{3}\). Find the work done by the force in first 4 \(s\).
355836
When a rubber-band is stretched by a distance \(x\), it exerts a restoring force of magnitude \(F=a x+b x^{2}\) where \(a\) and \(b\) are constants. The work done in stretching the unstrected rubber band by \(L\) is
355834
A 10 \(kg\) brick moves along an \(x\)-axis. Its acceleration as a function of its position is shown in figure. What is the net work performed on the brick by the force causing the acceleration as the brick moves from \(x=0\) to \(x = 8.0\,m\) ?
1 4\(J\)
2 8\(J\)
3 2\(J\)
4 1\(J\)
Explanation:
According to the graph the acceleration \(a\) varies linearly with the coordinate \(x\). We may write \(a=\alpha x\), where \(\alpha\) is the slope of the graph \(\alpha = \frac{{20}}{8} = 2.5\;m{s^{ - 1}}\) The force on the brick is in the positive \(x\) direction and according to Newton's second law, its magnitude is given by \(F=\dfrac{a}{m}=\dfrac{\alpha}{m} x\) if \(x_{f}\) is the final coordinate, the work done by the force is \(\begin{aligned}W & =\int_{0}^{x_{f}} F d x=\dfrac{\alpha}{m} \int_{0}^{x_{f}} x d x \\& =\dfrac{\alpha}{2 m} x_{f}^{2}=\dfrac{2.5}{2 \times 10} \times(8)^{2}=8 J\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355835
A particle of mass 6 \(kg\) moves according to the law \(x=0.2 t^{2}+0.02 t^{3}\). Find the work done by the force in first 4 \(s\).
355836
When a rubber-band is stretched by a distance \(x\), it exerts a restoring force of magnitude \(F=a x+b x^{2}\) where \(a\) and \(b\) are constants. The work done in stretching the unstrected rubber band by \(L\) is
355834
A 10 \(kg\) brick moves along an \(x\)-axis. Its acceleration as a function of its position is shown in figure. What is the net work performed on the brick by the force causing the acceleration as the brick moves from \(x=0\) to \(x = 8.0\,m\) ?
1 4\(J\)
2 8\(J\)
3 2\(J\)
4 1\(J\)
Explanation:
According to the graph the acceleration \(a\) varies linearly with the coordinate \(x\). We may write \(a=\alpha x\), where \(\alpha\) is the slope of the graph \(\alpha = \frac{{20}}{8} = 2.5\;m{s^{ - 1}}\) The force on the brick is in the positive \(x\) direction and according to Newton's second law, its magnitude is given by \(F=\dfrac{a}{m}=\dfrac{\alpha}{m} x\) if \(x_{f}\) is the final coordinate, the work done by the force is \(\begin{aligned}W & =\int_{0}^{x_{f}} F d x=\dfrac{\alpha}{m} \int_{0}^{x_{f}} x d x \\& =\dfrac{\alpha}{2 m} x_{f}^{2}=\dfrac{2.5}{2 \times 10} \times(8)^{2}=8 J\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355835
A particle of mass 6 \(kg\) moves according to the law \(x=0.2 t^{2}+0.02 t^{3}\). Find the work done by the force in first 4 \(s\).
355836
When a rubber-band is stretched by a distance \(x\), it exerts a restoring force of magnitude \(F=a x+b x^{2}\) where \(a\) and \(b\) are constants. The work done in stretching the unstrected rubber band by \(L\) is
