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

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.

$W = F_r*d \rightarrow F_r =$  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,

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2} (1) v^2$

$KE = 0.5v^2$

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,

$v= \sqrt{0.0441/0.5}$

$v= 0.9391m/s$