Question 22M.2.SL.TZ1.1b
| Date | May 2022 | Marks available | [Maximum mark: 2] | Reference code | 22M.2.SL.TZ1.1b |
| Level | SL | Paper | 2 | Time zone | TZ1 |
| Command term | Calculate | Question number | b | 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.
The vertical acceleration of the load downwards is 2.4 m s−2.
Calculate the tension in the string.
[2]
use of ma = mg − T «3.5 x 2.4 = 3.5g − T »
OR
T = 3.5(g − 2.4) ✓
26 «N» ✓
Accept 27 N from g = 10 m s−2
Some students simply calculated the net force acting on the load and did not recognize that this was not the tension force. Many set up a net force equation but had the direction of the forces backwards. This generally resulted from sloppy problem solving.
