Question 5

The velocity-time graph of a particle of mass 10 kg is shown in the figure. The net work done on the particle in the first two seconds of the motion is

$$W_{net} = \Delta K.E. = \frac{1}{2} m v^2 - \frac{1}{2} m u^2$$

$$W_{net} = \frac{1}{2} m (v^2 - u^2)$$

$$a = \frac{v_f - v_i}{t_f - t_i} = \frac{0 - 50}{10 - 0} = -5\text{ m/s}^2$$

$$u = 50\text{m/s}$$

$$v = u + at$$: $$v = 50 + (-5)(2)$$

$$W = \frac{1}{2} \times 10 \times (40^2 - 50^2)$$

$$W = 5 \times (1600 - 2500)$$

$$W = -4500\text{ J}$$

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