Moderate -0.5 This is a straightforward energy conservation problem requiring application of the work-energy principle. Students must calculate initial KE, final KE, and loss in PE, then find work done against resistance as the difference. It's a standard M1 question with clear given values and a direct method, requiring no novel insight—slightly easier than average due to its routine nature and minimal problem-solving demand.
A block of mass 15 kg slides down a line of greatest slope of an inclined plane. The top of the plane is at a vertical height of 1.6 m above the level of the bottom of the plane. The speed of the block at the top of the plane is 2 m s\(^{-1}\) and the speed of the block at the bottom of the plane is 4 m s\(^{-1}\).
Find the work done against the resistance to motion of the block. [4]
Question 1:
1 | 1 1
1522 =30 1542=120
2 2 | B1 | For KE at top or bottom.
Need not be evaluated.
1
15(4−2)2
is B0.
2
15g1.6 =240 | B1 | For PE change.
Need not be evaluated.
240+30=120+W | M1 | Attempt at work energy equation; 4
relevant terms; dimensionally
correct; allow sign errors.
1
15(4−2)2
is M0.
2
If W = F times a numerical distance
seen, then M0.
Work done = 150J | A1
Question | Answer | Marks | Guidance
1 | Alternative method for Q1
1.6
42 =22 +2a
sin | *M1 | Attempt to use v2 =u2 +2as with
1.6 1.6
s= or but not
sin cos
s=1.6sin or 1.6cos or 1.6.
If is given a value, then M0.
Must be using speeds 2 and 4 here.
15gsin−R=15a | DM1 | 3 terms; allow sign errors; allow
sin/cos mix but weight must be
resolved; dimensionally correct.
R=93.75sin | A1 | R=93.75cos
Must be consistent with theirs .
1.6
Work done = 93.75sin =150 J
sin | A1
4
Question | Answer | Marks | Guidance
A block of mass 15 kg slides down a line of greatest slope of an inclined plane. The top of the plane is at a vertical height of 1.6 m above the level of the bottom of the plane. The speed of the block at the top of the plane is 2 m s$^{-1}$ and the speed of the block at the bottom of the plane is 4 m s$^{-1}$.
Find the work done against the resistance to motion of the block. [4]
\hfill \mbox{\textit{CAIE M1 2023 Q1 [4]}}