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

0.225, 4.41 N

Step-by-step explanation:

A
`2\text{ }kg` wooden block has an initial speed of
`3\text{ }(m)/(s)`. It slides for
`2\text{ }m` before it comes to rest.

Let us assume the coefficient of friction to be
`\mu`. Hence, the friction force is given by
`\mu * N`, where
`N` is the Normal reaction from the ground.

`N\text{ = }m* g`

`m` is the mass of block and
`g` is the gravity. Let the displacement be
`S`.

`\text{Friction Force = }\mu * N\text{ = }\mu* m* g`

Work done by the frictional force = Kinetic Energy drained from the block

`\text{Work by friction = Friction force }*\text{Displacement = }\Delta KE\text{ = }(1)/(2)* m* v^(2)`

`\mu * m* g* S=(1)/(2)* m* v^(2)`

`\mu* 10(m)/(s^(2))* 2\text{ }m= (1)/(2)* (3(m)/(s) )^(2)`

`\mu=(9)/(40)=0.225`

Force required to keep the block moving at constant speed across the floor is equal to friction force =

`\mu* m* g=0.225*2\text{ }kg*9.8(m)/(s^(2))=4.41\text{ }N`.

∴ Coefficient of friction =
`0.225`

Force required to maintain constant speed =
`4.41N`

Answer 2

(a) The coefficient of friction is 0.153.

(b) Force needed to keep the block moving at a constant speed across the floor is 3 N in the direction opposite to that of frictional force.

Step-by-step explanation:

Given:

Mass of the block is,
`m=2\ kg`

Initial speed of the block is,
`u=3\ m/s`

Displacement of the block is,
`d=2\ m`

Final speed of the block is,
`v=0\ m/s`

(a)

Using Newton's equations of motion and determining the acceleration of the block:

`v^2=u^2+2ad\\0=3^2+2* a* 3\\0=9+6a\\6a=-9\\a=-(9)/(6)=-1.5\ m/s^2`

Therefore, frictional force acts on the block to cause the deceleration of the block.

From Newton's second law, net force is product of mass and acceleration.

Here, the only force acting along the motion of the block is friction.

So, friction acting on the block is given as:

`f=ma\\f=2* -1.5=-3\ N`

Negative sign shows the nature of frictional force that it acts opposite to the motion of the block.

Now, magnitude of frictional force is also given as:

`f=\mu N\\\mu =(f)/(N)`

where,
`\mu` is the coefficient of friction and
`N` is the normal force.

Here, there is no vertical motion. So, normal force is equal to weight of the block.

`\therefore N=mg=2* 9.8=19.6\ N`

`\therefore \mu=(f)/(N)=(3)/(19.6)=0.153`

Therefore, the coefficient of friction is 0.153.

(b)

In order to make the speed as constant, the net acceleration acting on the block must be 0. So, the forward force must be equal to the backward frictional force. Hence,

`\textrm{Force needed}=f=3\ N`

Therefore, force needed to keep the block moving at a constant speed across the floor is 3 N in the direction opposite to that of frictional force.