Question

After the collision, the block slides 1.00 m across the frictionless surface and an additional 0.450 m, before coming to rest, across a horizontal surface where the coefficient of friction between the block and the surface is 0.100. Use g = 9.80m/s^2. Calculate the speed the block has immediately after the collision .

Answer 1

v= 0.9391m/s

Step-by-step explanation:

We apply conservative energy equation, where all the work done by all forces is equal to change in Kinetic Energy.

`W = F_r*d \rightarrow F_r =` Frictional Force

`W= \mu N*d`

`W = \mu mgd`

`W = 0.1*9.8*0.45`

`W= 0.441J`

The change in Kinetic Energy is given by,

`KE = (1)/(2)mv^2`

`KE = (1)/(2) (1) v^2`

`KE = 0.5v^2`

How the work done by all force is equal to the change in KE, we have that

`W = KE`

`0.0441 = 0.5v^2`

Solving v,

`v= √(0.0441/0.5)`

`v= 0.9391m/s`