Standard +0.3 This is a straightforward work-energy problem requiring calculation of work done against gravity, resistance, and kinetic energy gained, then dividing by time (found from kinematics). All steps are standard A-level mechanics techniques with no novel insight required, making it slightly easier than average.
4 A particle \(B\) of mass 5 kg is at rest at the bottom of a slope which is angled at \(\sin ^ { - 1 } 0.2\) above the horizontal. A constant force \(D\) initially acts directly up the slope on \(B\).
The total resistance to the motion of \(B\) is modelled as being a constant 12 N .
At the instant that \(D\) stops acting, the speed of \(B\) is \(18 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and \(B\) has moved 90 m up the slope.
Determine the average power of \(D\) over the time that \(D\) has been acting on \(B\).
Acquired \(PE = 5 \times g \times 90 \times 0.2\) (J) \(= 90g\) or \(882\)
B1
AO 1.1; ignore wrong sign
\(90 = \frac{1}{2}(18+0)t \Rightarrow t = 10\)
M1
AO 3.1b; use of suvat to find time; or average speed \(= \frac{1}{2}(18+0)\)
Average power \(= (810 + 882 + 1080)\)/``\(10\)''
M1
AO 1.1; using Power \(= WD\)/time or Energy/dist \(\times\) average speed; allow 1 error e.g. missing energy term or sign error but not extra terms
\(= 277\) (W)
A1
AO 2.2a; \(277.2\)
Alternative method using \(P = Fv\)
Answer
Marks
Guidance
Answer
Marks
Guidance
Weight down slope \(= 5g \times 0.2 = 9.8\) N
B1
\(a = \frac{18^2}{2 \times 90} = 1.8\ \text{ms}^{-2}\); no other values of \(u\), \(v\) or \(s\) allowed
B1
Or finding \(KE\) (\(= 810\))
\(ma = 5 \times\) ``\(1.8\)'' \(= 9\) N
B1FT
Or \(ma = \frac{\text{``}810\text{''}}{90}\)
Average speed \(= \frac{1}{2}(18+0) = 9\) m/s
M1
Or \(P_{(\max)} = (12 +\) ``\(9.8\)'' \(+\) ``\(9\)''\() \times 18\) (\(= 554.4\)); or \(t = \frac{90}{\text{``}9\text{''}}\) (\(= 10s\)); allow 1 error e.g. missing energy term or sign error but not extra terms
Average power \(= (12 +\) ``\(9.8\)'' \(+\) ``\(9\)''\() \times\) ``\(9\)''
M1
Driving force \(\times\) average speed; or \(P_{av} = \frac{P_{\max}}{2} = \frac{\text{``}554.4\text{''}}{2}\); allow 1 error as per above
\(= 277\) (W)
A1
\(277.2\)
# Question 4:
**Method using $P = \frac{WD}{t}$**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Energy lost to resistance $= 12 \times 90$ (J) $= 1080$ | B1 | AO 1.1 |
| Acquired $KE = \frac{1}{2} \times 5 \times 18^2$ (J) $= 810$ | B1 | AO 3.1b |
| Acquired $PE = 5 \times g \times 90 \times 0.2$ (J) $= 90g$ or $882$ | B1 | AO 1.1; ignore wrong sign |
| $90 = \frac{1}{2}(18+0)t \Rightarrow t = 10$ | M1 | AO 3.1b; use of suvat to find time; or average speed $= \frac{1}{2}(18+0)$ |
| Average power $= (810 + 882 + 1080)$/``$10$'' | M1 | AO 1.1; using Power $= WD$/time or Energy/dist $\times$ average speed; allow 1 error e.g. missing energy term or sign error but not extra terms |
| $= 277$ (W) | A1 | AO 2.2a; $277.2$ |
**Alternative method using $P = Fv$**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Weight down slope $= 5g \times 0.2 = 9.8$ N | B1 | |
| $a = \frac{18^2}{2 \times 90} = 1.8\ \text{ms}^{-2}$; no other values of $u$, $v$ or $s$ allowed | B1 | Or finding $KE$ ($= 810$) |
| $ma = 5 \times$ ``$1.8$'' $= 9$ N | B1FT | Or $ma = \frac{\text{``}810\text{''}}{90}$ |
| Average speed $= \frac{1}{2}(18+0) = 9$ m/s | M1 | Or $P_{(\max)} = (12 +$ ``$9.8$'' $+$ ``$9$''$) \times 18$ ($= 554.4$); or $t = \frac{90}{\text{``}9\text{''}}$ ($= 10s$); allow 1 error e.g. missing energy term or sign error but not extra terms |
| Average power $= (12 +$ ``$9.8$'' $+$ ``$9$''$) \times$ ``$9$'' | M1 | Driving force $\times$ average speed; or $P_{av} = \frac{P_{\max}}{2} = \frac{\text{``}554.4\text{''}}{2}$; allow 1 error as per above |
| $= 277$ (W) | A1 | $277.2$ |
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4 A particle $B$ of mass 5 kg is at rest at the bottom of a slope which is angled at $\sin ^ { - 1 } 0.2$ above the horizontal. A constant force $D$ initially acts directly up the slope on $B$.
The total resistance to the motion of $B$ is modelled as being a constant 12 N .\\
At the instant that $D$ stops acting, the speed of $B$ is $18 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $B$ has moved 90 m up the slope.\\
Determine the average power of $D$ over the time that $D$ has been acting on $B$.
\hfill \mbox{\textit{OCR Further Mechanics AS 2024 Q4 [6]}}