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 .

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cjmejiab cjmejiab · Answer 1 · 2019-11-07T23:56:13+01:00

Answer:

v= 0.9391m/s

Explanation:

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

[tex]W = F_r*d \rightarrow F_r =[/tex] Frictional Force

[tex]W= \mu N*d[/tex]

[tex]W = \mu mgd[/tex]

[tex]W = 0.1*9.8*0.45[/tex]

[tex]W= 0.441J[/tex]

The change in Kinetic Energy is given by,

[tex]KE = \frac{1}{2}mv^2[/tex]

[tex]KE = \frac{1}{2} (1) v^2[/tex]

[tex]KE = 0.5v^2[/tex]

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

[tex]W = KE[/tex]

[tex]0.0441 = 0.5v^2[/tex]

Solving v,

[tex]v= \sqrt{0.0441/0.5}[/tex]

[tex]v= 0.9391m/s[/tex]