Question #1c2bd
1 Answer
Explanation:
Before the force is applied we know
"momentum" = "mass" times "velocity" = "10 kg m/s"momentum=mass×velocity=10 kg m/s
Since mass in the above is
We can now work out kinetic energy, KE:
KE=1/2mv^2=1/2*5*2^2="10 J"KE=12mv2=12⋅5⋅22=10 J
The force of
F=(mv-"mu")/t=m * (v-u)/tF=mv−mut=m⋅v−ut
Where
So
0.2 = 5 * ((v-2))/100.2=5⋅(v−2)10
rearranging helps us find
2 = 5 * (v-2)2=5⋅(v−2)
2/5 = v-225=v−2
v = 12/5" ms"^(-1)v=125 ms−1
Hence final kinetic energy,
KE = 1/2mv^2 = 1/2 * 5 * (12/5)^2 = "14.4 J"KE=12mv2=12⋅5⋅(125)2=14.4 J
So the overall change in
Delta_"KE" = 14.4 - 10 = "4.4 J"