AP Physics C Mechanics Conservation of Energy FRQ

(a) For correctly using conservation of energy \(U_1+K_1=U_2+K_2\) \(U_1+0=U_2+\frac{1}{2}mv^2\) For correctly substituting values from the graph into the equation above

\((8j)=(6j)+\left ( \frac{1}{2} \right )(0.40kg)v^2\)

\(v=3.16m/s\) 

(b)
i) For correctly relating the slope of the line to the force exerted on the particle \(F=-slope=-\frac{(U_2-U_1)}{(X_2-X_1)}=-\frac{(12-0)J}{(0-6)M}=2.0N\)

For using Newton’s second law to calculate the acceleration of the particle \(a=\frac{F}{m}=\frac{(2.0N)}{(0.40kg)}=5.0/s^2\)

Note: point is awarded for \(5.0 m/ s^2\)

(b) 
i) Alternate Solution 

Solving for two velocities between 2.0 m and 6.0 m\( v_1 = 0\)

ii) For selecting answer consistent with part (b)(i)

(c) For a correct answer \(F=-slope=0\therefore a=0\) Note: credit is given even if no work is shown

(d)

i) For using a correct kinematic equation to calculate the time

\(d=v_1t+\frac{1}{2}at^2=0+\frac{1}{2}at^2\)

For correctly substituting acceleration from part (b)(i) into equation above

\(t=\sqrt{\frac{2d}{a}}=\frac{(2)(6.0-2.0)m}{(5.0m/s^2)}=1.26s\)

ii) For using a correct kinematic equation to determine the speed at x = 6.0 m or using the velocity at 6.0 m from alternate solutions part (b)(i) or (d)(i)

\(v^2_2=v^2_1+2ad=0+2ad
v_2=\sqrt{2ad}=\sqrt{(2)(5.0m/s^2)(6.0-2.0)m} v_{2}=6.32m/s

For using a kinematic equation for zero acceleration to calculate the time d =vt 
\(t=\frac{d}{v}=\frac{(8.0-6.0)m}{(6.32m/s)}\) t=0.32s

Alternate Solution 

For correctly using the energy equation to solve for the speed at 6.0 m (or using the value from alternate solution part (b)(i)

\(U_{1}+K_{1}=U_{2}+K_{2} \therefore (8J) +0+\frac{1}{2}(0.40kg)v^{2}_{2} v_{2}=6.32m/s\)

For using a kinematic equation with zero acceleration to calculate the time d=vt\( t=d/v=(8.0-6.0)m/(6.32m/s) t=0.32s\)

(d) 
iii) For using a correct kinematic equation to calculate the time

\(d=\frac{1}{2}(v_1+v_2)t\)
\(t=\frac{2d}{(v_1+v_2)}=\frac{(2)(14-8.0)m}{(6.32m/s+3.16m/s)} t=1.26s\)

(e) For correctly using conservation of energy to calculate the speed

\(U_1+K_1=U_2+K_2\)
\(0+\frac{1}{2}m_1v^2_1\)
\(0+\frac{1}{2}m_1v^2_1=U_2+\frac{1}{2}m_2v^2_2\)
\(\left ( \frac{1}{2} \right )(0.40kg)v^2_1=(6j)+\left ( \frac{1}{2} \right )(0.40kg)(1.0m/s^2)
v=5.57m/s\)

(f) For a horizontal line at F = 2 N from x = 0 m to x = 6 m 
For a horizontal line at F = 0 from x = 6 m to x = 8 m and from x = 14 m to x = 16 m 
For a horizontal line at F = –1 N from x = 8 m to x = 14 m
Note: Two points earned if shape and magnitudes are correct but graph is inverted