Potential Energy AP Physics 1 FRQ – Exam Style Questions etc.
Potential Energy AP Physics 1 FRQ
Unit: 3. Work , Energy and Power
Weightage : 10-15%
Exam Style Practice Questions, Potential Energy AP Physics 1 FRQ
Question
An ideal spring of unstretched length 0.20 m is placed horizontally on a frictionless table as shown above. One end of the spring is fixed and the other end is attached to a block of mass M = 8.0 kg. The 8.0 kg block is also attached to a massless string that passes over a small frictionless pulley. A block of mass m = 4.0 kg hangs from the other end of the string. When this spring-and-blocks system is in equilibrium, the length of the spring is 0.25 m and the 4.0 kg block is 0.70 m above the floor.
(a) On the figures below, draw free-body diagrams showing and labeling the forces on each block when the system is in equilibrium.
M = 8.0 kg m = 4.0 kg
(b) Calculate the tension in the string.
(c) Calculate the force constant of the spring.
The string is now cut at point P.
(d) Calculate the time taken by the 4.0 kg block to hit the floor.
(e) Calculate the maximum speed attained by the 8.0 kg block as it oscillates back and forth
Answer/Explanation
Ans:
(a) FBD 
(b) Simply isolating the 4 kg mass at rest. Fnet = 0 Ft – mg = 0 Ft = 39 N
(c) Tension in string is uniform throughout, now looking at the 8 kg mass,
Fsp = Ft = k∆x 39 = k (0.05) k = 780 N/m
(d) 4 kg mass is in free fall. D = vit + ½ g t2 – 0.7 = 0 + ½ (– 9.8)t2 t = 0.38 sec
(e) The 8 kg block will be pulled towards the wall and will reach a maximum speed when it passes the relaxed
length of the spring. At this point all of the initial stored potential energy is converted to kinetic energy
Usp = K ½ k ∆x2 = ½ mv2 ½ (780) (0.05) = ½ (8) v2 v = 0.49 m/s
Question
As shown above, a 0.20-kilogram mass is sliding on a horizontal, frictionless air track with a speed of 3.0 meters per second when it instantaneously hits and sticks to a 1.3-kilogram mass initially at rest on the track. The 1.3-kilogram mass is connected to one end of a massless spring, which has a spring constant of 100 newtons per meter. The other end of the spring is fixed.
a. Determine the following for the 0.20-kilogram mass immediately before the impact.
i. Its linear momentum ii. Its kinetic energy
b. Determine the following for the combined masses immediately after the impact.
i. The linear momentum ii. The kinetic energy
After the collision, the two masses compress the spring as shown.
c. Determine the maximum compression distance of the spring.
▶️Answer/Explanation
Ans:
a) i) p = mv = (0.2)(3) = 0.6 kg m/s
ii) K = ½ mv2 = ½ (0.2)(3)2 = 0.9 J
b) i.) Apply momentum conservation pbefore = pafter = 0.6 kg m/s
ii) First find the velocity after using the momentum above
0.6 = (1.3+0.2) vf vf = 0.4 m/s K = ½ (m1+m2) vf2 = ½ (1.3+0.2)(0.4)2 = 0.12 J
c) Apply energy conservation K = Usp 0.12 J = ½ k∆x2 = ½ (100) ∆x2 ∆x = 0.05 m