Question

the figure below shows an object with a mass of m = 5.70 kg that starts from rest at point a and slides on a track with negligible friction. point a is at a height of h_a = 5.30 m. (a) what is the objects speed at point b (in m/s)? m/s what is the objects speed at point c (in m/s)? m/s (b) what is the net work (in j) done by the gravitational force on the object as it moves from point a to point c? j

Explanation:

Step1: Apply conservation of mechanical energy

The initial mechanical energy at point A is $E_{A}=mgh_{A}$ (potential energy only as it starts from rest, $v_{A} = 0$). At point B, the mechanical energy is $E_{B}=mgh_{B}+\frac{1}{2}mv_{B}^{2}$. Since there is no friction, $E_{A}=E_{B}$. So $mgh_{A}=mgh_{B}+\frac{1}{2}mv_{B}^{2}$. We can cancel out the mass $m$ from both sides of the equation: $gh_{A}=gh_{B}+\frac{1}{2}v_{B}^{2}$.

Step2: Solve for $v_{B}$

We know that $g = 9.8\ m/s^{2}$, $h_{A}=5.30\ m$ and $h_{B}=3.20\ m$. Rearranging the equation $gh_{A}=gh_{B}+\frac{1}{2}v_{B}^{2}$ for $v_{B}$ gives $v_{B}=\sqrt{2g(h_{A}-h_{B})}$. Substituting the values: $v_{B}=\sqrt{2\times9.8\times(5.30 - 3.20)}=\sqrt{2\times9.8\times2.1}=\sqrt{41.16}\approx6.41\ m/s$.

Step3: Apply conservation of mechanical energy for point C

At point A, $E_{A}=mgh_{A}$ and at point C, $E_{C}=mgh_{C}+\frac{1}{2}mv_{C}^{2}$. Since $E_{A}=E_{C}$ and canceling out $m$, we have $gh_{A}=gh_{C}+\frac{1}{2}v_{C}^{2}$. Rearranging for $v_{C}$ gives $v_{C}=\sqrt{2g(h_{A}-h_{C})}$. Given $h_{C}=2.00\ m$, $v_{C}=\sqrt{2\times9.8\times(5.30 - 2.00)}=\sqrt{2\times9.8\times3.3}=\sqrt{64.68}\approx8.04\ m/s$.

Step4: Calculate the work - done by gravitational force

The work - done by the gravitational force $W_{g}$ when the object moves from point A to point C is given by $W_{g}=mg\Delta h$, where $\Delta h=h_{A}-h_{C}$. Substituting $m = 5.70\ kg$, $g = 9.8\ m/s^{2}$, $h_{A}=5.30\ m$ and $h_{C}=2.00\ m$, we get $W_{g}=5.70\times9.8\times(5.30 - 2.00)=5.70\times9.8\times3.3 = 185.982\approx186\ J$.

Answer:

a. $v_{B}\approx6.41\ m/s$, $v_{C}\approx8.04\ m/s$
b. $186\ J$