Work, Energy and Power: Question 7

Worked solution

Part (a): Work done by the applied force

Work done is force multiplied by the displacement in the direction of the force. The applied force and the displacement are both horizontal, so: $W_{\text{applied}} = Fs = 90\times6.0 = 540\text{ J}$

Check: $90\times6.0=90\times6=540\text{ J}$. Confirmed.

Part (b): Work done against friction

Friction acts along the same line as the displacement (opposing it), so the work done against it is also found from $W=Fs$: $W_{\text{friction}} = 30\times6.0 = 180\text{ J}$

Check: $30\times6.0=180\text{ J}$. Confirmed.

Part (c): Kinetic energy gained (work–energy principle)

The work–energy principle states that the net (resultant) work done on an object equals its change in kinetic energy. The applied force does positive work, while friction removes energy, so: $E_k = W_{\text{applied}} - W_{\text{friction}} = 540 - 180 = 360\text{ J}$

Since the crate starts from rest, this net work equals the kinetic energy gained, which is also the crate’s final kinetic energy.

Check: $540-180=360\text{ J}$. Confirmed.

Part (d): Final speed of the crate

Using $E_k=\tfrac12mv^2$, rearranged for $v$: $v = \sqrt{\frac{2E_k}{m}} = \sqrt{\frac{2\times360}{50}}$

Working the numerator first: $2\times360=720$. Dividing by the mass: $720/50=14.4$.

$v = \sqrt{14.4} \approx 3.79\text{ m s}^{-1}$

Check: $3.79^2=14.3641$ and $3.80^2=14.44$, both close to $14.4$, confirming $v\approx3.79\text{ m s}^{-1}$ (3 s.f.).

Final answers

• (a) Work done by the applied force $= \boxed{540}\text{ J}$
• (b) Work done against friction $= \boxed{180}\text{ J}$
• (c) Kinetic energy gained $= \boxed{360}\text{ J}$
• (d) Final speed $\approx \boxed{3.79}\text{ m s}^{-1}$