An object of weight 800 N is pulled up a plane inclined at 15° using an effort of 750 N as shown in figure Q1a. Determine the coefficient of friction between the object and the plane.

The normal force ($N$) acts perpendicular to the plane, balancing the perpendicular component of the weight ($W \cos \theta$).

$$N = W \cos \theta$$ $$N = 800 \, \text{N} \times \cos 45^\circ$$ $$N = 800 \, \text{N} \times \frac{\sqrt{2}}{2}$$ $$N = 400 \sqrt{2} \, \text{N} \approx 565.69 \, \text{N}$$

Since the object is being pulled up the plane, the applied force ($P$) acts upwards, while the component of weight parallel to the plane ($W \sin \theta$) and the frictional force ($F_f$) act downwards. Assuming constant velocity or impending motion, the net force is zero.

$$P = W \sin \theta + F_f$$

We know that the frictional force is given by $F_f = \mu N$, where $\mu$ is the coefficient of friction.

$$P = W \sin \theta + \mu N$$

$$\mu (W \cos \theta) = P - W \sin \theta$$ $$\mu = \frac{P - W \sin \theta}{W \cos \theta}$$

Now, substitute the given values:

$$\mu = \frac{750 \, \text{N} - (800 \, \text{N} \times \sin 45^\circ)}{800 \, \text{N} \times \cos 45^\circ}$$ $$\mu = \frac{750 - (800 \times \frac{\sqrt{2}}{2})}{800 \times \frac{\sqrt{2}}{2}}$$ $$\mu = \frac{750 - 400 \sqrt{2}}{400 \sqrt{2}}$$ $$\mu = \frac{750}{400 \sqrt{2}} - 1$$ $$\mu = \frac{15}{8 \sqrt{2}} - 1$$ $$\mu = \frac{15 \sqrt{2}}{16} - 1$$ $$\mu \approx 1.3258 - 1$$ $$\mu \approx 0.3258$$

Rounding to three significant figures:

$$\mu \approx 0.326$$

The coefficient of friction between the object and the plane is $\boxed{0.326}$.