The correct answer is C because:

3-6-mcq-m-ma4-ocr-al-physics

Method 1: Work Done

Find the work done by the 5.0 N force:
- Work done = F cosθ
- Work done = 5 x 12 x cos 30
- Work done = 52.0 J
Find the kinetic energy gained by the object after it has moved 12m:
- Transfer of energy = work done.
- Since no other energy transfers are taking place, gain in kinetic energy = work done.
- Gain in kinetic energy = 52.0 J

A is incorrect as the mass has not been included when calculating $\frac{1}{2} m v^{2}$

B is incorrect as when resolving the force in the work done equation, sin θ rather than cos θ has been used.

D is incorrect as the $\frac{1}{2}$ has not been included when calculating $\frac{1}{2} m v^{2}$

Alternative method: Motion Equations

Calculate acceleration, a, in the direction of motion:
- $a = \frac{F}{m} = \frac{5 c o s 30}{4}$ = 1.08 ms^-2
Calculate the speed of the object after moving 12 m:
- $v^{2} = u^{2} + 2 a s$
- $v^{2} = 0^{2} + 2 \times 1.08 \times 12$
- $v^{2} = 25.98$
- v = 5.10 ms^-1
Calculate the kinetic energy of the object after moving 12 m:
- $E_{k} = \frac{1}{2} m v^{2} = \frac{1}{2} \times 4 \times 5 . 10^{2} = 52.0 J$

The second method is not as quick as the first one, and it is important to note that if the object had not been moving in a straight line, the second method would not be valid.
- This shows the importance of understanding the principle: work done = energy transferred. It is extremely useful.

Work, Energy & Power (OCR AS Physics)

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