Question

A 60kg block innitially at rest is pulled to the right a long a horizontal force of12N. Find the speed of the block after it has moved 3M if the surface in contact have a coefficient of kinetic friction of 0.15​

Answer 1

Answer:

The speed of the block after it has moved 3M if the surface in contact have a coefficient of kinetic friction of 0.15​  is 1.7s m/sec

Explanation:

Given:

mass of the block = 6.0 kg

Force with which the block is pulled = 12 N

Kinetic friction \mu= 0.15

Distance travelled  s = 3 m

To Find:

speed of the block after it has moved 3 metres =?

Solution :

W know that the friction formula is

$f_k = \mu m g$

Substituting the values,

$f_k = (0.15)(6)(10)$

$f_k= 9 N$

Now Acceleration is Given by

$a=\frac{F -f_k}{m}$

$a=\frac{12 - 9}{60}$

$a=\frac{3}{6}$

a= 0.5 $m/s^2$

Initial velocity is u = 0

Also we know that,

$v^2 - u^2=2as$

So the equation becomes

$v^2 =2as$

$v=\sqrt{2as}$

Substituting the values,

$v=\sqrt{2as}$

$v=\sqrt{2(0.5)(3)}$

$v= \sqrt{3}$

v= 1.73 m/s