Question

3. A 2kg wooden block whose initial speed is 3 m/s slides on a smooth floor for 2 meters before it comes to a
stop.
a. Find the coefficient of friction?
b. How much force would be needed to keep the block moving at a constant speed across the floor?

Answer 1

Answer:

Calculating Coefficient of friction is 0.229.

Force is 4.5 N that keep the block moving at a constant speed.

Explanation:

We know that speed expression is as $\mathrm{V}^{2}=\mathrm{V}_{\mathrm{i}}^{2}+2 . \mathrm{a} . \Delta \mathrm{s}$.

Where, ${V}_{i}$ is initial speed, V is final speed, ∆s displacement and a acceleration.

Given that,

${V}_{i}$ =3 m/s, V = 0 m/s, and  ∆s = 2 m

Substitute the values in the above formula,

$0=3^{2}-2 \times 2 \times a$

0 = 9 - 4a

4a = 9

$a=2.25 \mathrm{m} / \mathrm{s}^{2}$

$a=2.25 \mathrm{m} / \mathrm{s}^{2}$ is the acceleration.

Calculating Coefficient of friction:

$\mathrm{F}=\mathrm{m} \times \mathrm{a}$

$\mathrm{F}=\mu \times \mathrm{m} \times \mathrm{g}$

Compare the above equation

$\mu \times m \times g=m \times a$

Cancel "m" common term in both L.H.S and R.H.S

$\text { Equation becomes, } \mu \times g=a$

$\text { Coefficient of friction } \mu=\frac{a}{g}$

$\mathrm{g} \text { on earth surface }=9.8 \mathrm{m} / \mathrm{s}^{2}$

$\mu=\frac{2.25}{9.8}$

$\mu=0.229$

Hence coefficient of friction is 0.229.

calculating force:

$\text { We know that } \mathrm{F}=\mathrm{m} \times \mathrm{a}$

$\mathrm{F}=2 \times 2.25 \quad(\mathrm{m}=2 \mathrm{kg} \text { given })$

F = 4.5 N

Therefore, the force would be 4.5 N to keep the block moving at a constant speed across the floor.