Plz help me with physics highschool?

Question

An 8.0 kg wood block is sliding along on a concrete floor at 15m/s. After 12 m its speed has been reduced to 5.0m/s by friction. How much work was done by

friction over the 12 m distance?

Answer 1

Work  W equal to force F times distance d that force was applied

W=Fd in this case the force is the force of friction f and still

f= ma

d= Vo t + 0.5 a t^2

V= Vo - at

t= (Vo - V)/ a

d= Vo (Vo - V)/ a + 0.5 [(Vo - V)/ a]^2

d a ^2  - Vo (Vo - V) a  - 0.5 [(Vo - V)]^2=0

Vo =15m/s

V= 5.0 m/s

d= 12 m

Now solve for a  and then

Work= m a d :)  where m = 8.0 kg

Answer 2

friction caused it to slow down thats all i can say

Answer 3

work = change in kinetic energy = 1/2 m (v_f)^2 - 1/2 m (v_i)^2

= 1/2 (8) (5^2 - 15^2) J

= -800J

Easy as that!

Answer 4

Mass m = 8.0 kg

Initial velocity u = 15 m/s

Final velocity v = 5 m/s

Displacement s = 12 m

Let a = acceleration

v^2 = u^2 + 2as

Or, a = (v^2 - u^2)/(2s)

Force of friction F = m*(v^2 - u^2)/(2s)

Work done = Fs = m*(v^2 - u^2)/2

= 8*(5^2 - 15^2)/2

= 8 * (25 - 225)/2

= 8 * (-200)/2

= -800 Joule

Ans: -800 Joule