Standard +0.3 This is a straightforward work-energy problem requiring resolution of forces and application of the work-energy principle. Students must calculate work done by applied force, work against gravity, and work against resistance, then use conservation of energy. While it involves multiple steps and careful angle resolution, it follows a standard M1 template with no novel insight required, making it slightly easier than average.
\includegraphics{figure_3}
A block of mass 10 kg is at rest on a rough plane inclined at an angle of 30° to the horizontal. A force of 120 N is applied to the block at an angle of 20° above a line of greatest slope (see diagram). There is a force resisting the motion of the block and 200 J of work is done against this force when the block has moved a distance of 5 m up the plane from rest.
Find the speed of the block when it has moved a distance of 5 m up the plane from rest. [5]
Work done by 120 N force = 1205cos20 =563.81557.
B1
(PE change=)10g5sin30 =250
B1
For PE change.
Attempt at work energy equation
M1
4 relevant terms; dimensionally
correct; allow sign errors; allow
sin/cos mix in relevant resolved
terms.
1
1205cos20−10g5sin30−200= 10v2
2
563.815−250−200=5v2
Answer
Marks
Guidance
A1
Speed = 4.77 m s–1
A1
awrt 4.77.
Question
Answer
Marks
3
Alternative method for Question 3
200
=40
Resistive force =
Answer
Marks
Guidance
5
*B1
oe e.g.5RF=200.
120 cos20−RF−10gsin30=10a
*M1
4 relevant terms; dimensionally
correct; allow sign errors; allow
sin/cos mix; allow with their
resistive force or just RF.
Answer
Marks
Guidance
a = 2.276…
A1
Allow arwt 2.3 to 2sf from correct
work.
Answer
Marks
Guidance
v2 =0+2(2.276)5
DM1
Use of v2 = u2 + 2as using u=0,
s=5 and their positive a which
has come from a resistive force
using work done.
Answer
Marks
Guidance
Speed = 4.77 m s–1
A1
awrt 4.77.
5
Answer
Marks
Guidance
Question
Answer
Marks
Question 3:
3 | Work done by 120 N force = 1205cos20 =563.81557. | B1
(PE change=)10g5sin30 =250 | B1 | For PE change.
Attempt at work energy equation | M1 | 4 relevant terms; dimensionally
correct; allow sign errors; allow
sin/cos mix in relevant resolved
terms.
1
1205cos20−10g5sin30−200= 10v2
2
563.815−250−200=5v2
| A1
Speed = 4.77 m s–1 | A1 | awrt 4.77.
Question | Answer | Marks | Guidance
3 | Alternative method for Question 3
200
=40
Resistive force =
5 | *B1 | oe e.g.5RF=200.
120 cos20−RF−10gsin30=10a | *M1 | 4 relevant terms; dimensionally
correct; allow sign errors; allow
sin/cos mix; allow with their
resistive force or just RF.
a = 2.276… | A1 | Allow arwt 2.3 to 2sf from correct
work.
v2 =0+2(2.276)5 | DM1 | Use of v2 = u2 + 2as using u=0,
s=5 and their positive a which
has come from a resistive force
using work done.
Speed = 4.77 m s–1 | A1 | awrt 4.77.
5
Question | Answer | Marks | Guidance
\includegraphics{figure_3}
A block of mass 10 kg is at rest on a rough plane inclined at an angle of 30° to the horizontal. A force of 120 N is applied to the block at an angle of 20° above a line of greatest slope (see diagram). There is a force resisting the motion of the block and 200 J of work is done against this force when the block has moved a distance of 5 m up the plane from rest.
Find the speed of the block when it has moved a distance of 5 m up the plane from rest. [5]
\hfill \mbox{\textit{CAIE M1 2023 Q3 [5]}}