Question

You take the same 3 kg metal block and slide it along the floor, where the coefficient of friction is only 0.4. You release the block with an initial velocity of 6 m/s. How long will it take for the block to come to stop? How far does the block move?

Answer 1

Answer:

1.52905 seconds

4.58715 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

$\mu$ = Coefficient of friction = 0.4

g = Acceleration due to gravity = 9.81 m/s²

a = Acceleration = $\mu g$

Equation of motion

$v=u+at\\\Rightarrow t=\frac{v-u}{\mu g}\\\Rightarrow t=\frac{0-6}{-0.4\times 9.81}\\\Rightarrow t=1.52905\ s$

It will take 1.52905 seconds for the block to slow down

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2\mu g}\\\Rightarrow s=\frac{0^2-6^2}{2\times 0.4\times -9.81}\\\Rightarrow s=4.58715\ m$

The block will travel 4.58715 m before it stops