Explanation:
F net of sled = Tension force by rope - Kinetic friction between ground.
F normal of sled = mg = (67kg)(9.81kg/m^2) = 657.27N.
Kinetic friction = 0.18 (I cannot see the value) * Normal force of sled = 0.18 * 657.27N = 118.31N
So F net of sled = 800N - 118.31N = 681.69N.
(I cannot see what the question is asking for, please check on your own!)
Acceleration a=3m/s^2
time t= 4.1seconds
Final velocity V= 55km/h
initial velocity U= ?
First convert V to m/s
36km/h=10m/s
55km/h= 55*10/36=15.28m/s
Using the formula V= U+at
U= V-at
U= 15.28-3*4.1=15.28-12.3=2.98m/s
Initial velocity U= 2.98m/s or 10.73km/h (Using the conversion rate 36km/h=10m/s)
c.charge due to the reaction process between the two
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Since D=M/V, the answer would be 2.7