More practice questions

  1. Q4. A porter pushes a cart with a force of 80 N for the first 6 m. Tiring, his push then falls linearly with position to 40 N over the next 4 m. A friction force of 40 N opposes the motion throughout. Find the work done by the porter and by friction over the 10 m.

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    The porter’s graph is a rectangle and a trapezium:

    WP=80×6+12(80+40)×4=480+240=720 J\begin{aligned} W_P &= 80 \times 6 + \tfrac{1}{2}(80 + 40) \times 4 \\ &= 480 + 240 = 720\ \text{J} \end{aligned}

    Friction is a rectangle below the axis: Wf = −40 × 10, that is −400 J.

  2. Q5. A spring of spring constant 400 N m is stretched slowly from its natural length by 5 cm. How much work is done on it?

    Show answerHide answer W=12kx2=12(400)(0.05)2=0.5 J\begin{aligned} W &= \tfrac{1}{2}kx^{2} \\ &= \tfrac{1}{2}(400)(0.05)^{2} \\ &= 0.5\ \text{J} \end{aligned}

  3. Q6. A force F = (4x − x2) N acts along the x-axis. Find the work done from x = 0 to x = 6 m. In which part of the motion is the work negative?

    Show answerHide answer W=∫06(4x−x2) dx=[2x2−13x3]06=72−72=0\begin{aligned} W &= \int_0^{6} (4x - x^{2})\,dx \\ &= \Big[2x^{2} - \tfrac{1}{3}x^{3}\Big]_0^{6} \\ &= 72 - 72 = 0 \end{aligned}

    The force is positive for 0 < x < 4 m and negative for 4 < x < 6 m, so the work is negative in the last 2 m. The positive and negative areas are equal (32/3 J each), so the total work is zero.