A 75 kg bobsled is pushed along a horizontal surface by two athlets. After the bobsled is pushed a distance?

Question

of 4.5 meters starting from rest, its speed is 6 m/s. Find the magnitude of the net force on the bobsled.

Answer 1

F=ma

a=change in velocity over time DV/t


Change of velocity is DV=V2-V1=6 m/s

also
S=a(t^2)/2 so a=2S/t^2

DV/t=2S/t^2
or t=2S/DV

a= DV/(2S/DV)=(DV)^2/(2S)
F=ma=
F=75(6)^2/(2 x 4.5)=300 Newtons

Answer 2

Given: Mass = 75kg Vo = 0 m/s V1 = 6 m/s D = 4.5m hit upon: The stress properly acceptable formula: F = m*a, mass is given, yet now no longer acceleration. First, hit upon acceleration. A = (V1-V0)/2t, t = d/V1 = 4.5/(2*6) = a million.5sec A = (6-0)/a million.5 = 4m/s^2 F = seventy 5*4 = 3 hundred kg*m/s^2

Answer 3

Given: Mass = 75kg
Vo = 0 m/s
V1 = 6 m/s
D = 4.5m

Find: The Force

Relevant formulas: F = m*a, mass is given, but not acceleration. First, find acceleration.

A = (V1-V0)/2t, t = d/V1 = 4.5/(2*6) = 1.5sec
A = (6-0)/1.5 = 4m/s^2

F = 75*4 = 300 kg*m/s^2