Question 22M.2.SL.TZ1.d
| Date | May 2022 | Marks available | [Maximum mark: 4] | Reference code | 22M.2.SL.TZ1.d |
| Level | SL | Paper | 2 | Time zone | TZ1 |
| Command term | Determine | Question number | d | Adapted from | N/A |
A student uses a load to pull a box up a ramp inclined at 30°. A string of constant length and negligible mass connects the box to the load that falls vertically. The string passes over a pulley that runs on a frictionless axle. Friction acts between the base of the box and the ramp. Air resistance is negligible.
The load has a mass of 3.5 kg and is initially 0.95 m above the floor. The mass of the box is 1.5 kg.
The load is released and accelerates downwards.
After the load has hit the floor, the box travels a further 0.35 m along the ramp before coming to rest. Determine the average frictional force between the box and the surface of the ramp.
[4]
ALTERNATIVE 1
«» leading to a = 6.3 «m s-2»
OR
« » leading to t = 0.33 « s » ✓
Fnet = « = » 9.45 «N» ✓
Weight down ramp = 1.5 x 9.8 x sin(30) = 7.4 «N» ✓
friction force = net force – weight down ramp = 2.1 «N» ✓
ALTERNATIVE 2
kinetic energy initial = work done to stop 0.5 x 1.5 x (2.1)2 = FNET x 0.35 ✓
Fnet = 9.45 «N» ✓
Weight down ramp = 1.5 x 9.8 x sin(30) = 7.4 «N» ✓
friction force = net force – weight down ramp = 2.1 «N» ✓
Accept 1.95 N from g = 10 m s-2.
Accept 2.42 N from u = 2.14 m s-1.
Candidates struggled with this question. Very few drew a clear free-body diagram and many simply calculated the acceleration of the box from the given information and used this to calculate the net force on the box, confusing this with the frictional force.
