A block takes time t to slide down a plane inclined at 45° to the horizontal; if the plane is made smooth (frictionless) it takes t/√2. Find the coefficient of friction between the block and the plane.

What the time change tells you

The block slides the same distance down the same incline in both cases, starting from rest. For constant acceleration, the only way the time can change is if the acceleration changes.

Write the distance relation for each case

From rest with constant acceleration, $$s=\frac{1}{2}at^2.$$

• With friction (time $t$, acceleration $a$): $$s=\frac{1}{2}at^2.$$
• Frictionless (time $t/\sqrt{2}$, acceleration $a_0$): $$s=\frac{1}{2}a_0\left(\frac{t}{\sqrt{2}}\right)^2=\frac{1}{2}a_0\cdot\frac{t^2}{2}=\frac{1}{4}a_0 t^2.$$

Equate distances: $$\frac{1}{2}at^2=\frac{1}{4}a_0 t^2 \;\Rightarrow\; a=\frac{a_0}{2}.$$

Express accelerations on a rough and smooth incline

For an incline of angle $\theta$:

• Frictionless acceleration: $$a_0=g\sin\theta.$$
• With kinetic friction coefficient $\mu$: $$a=g\left(\sin\theta-\mu\cos\theta\right).$$

Use $a=a_0/2$: $$g(\sin\theta-\mu\cos\theta)=\frac{1}{2}g\sin\theta.$$ Cancel $g$ and solve: $$\sin\theta-\mu\cos\theta=\frac{1}{2}\sin\theta$$ $$\mu\cos\theta=\frac{1}{2}\sin\theta$$ $$\mu=\frac{1}{2}\tan\theta.$$

Substitute $\theta=45^\circ$

$$\mu=\frac{1}{2}\tan 45^\circ=\frac{1}{2}(1)=0.5.$$