Question

A 75.0kg bobsled is pushed along a horizontal surface by two athletes. After the bobsled is pushed a distance of 4.5m starting from the rest, its speed is 6.0m/s. Find the magnitude of the net force on the bobsled.

Answer 1

a= v²/2s
F=ma

Answer 2

Fd=0.5mv^2
F(4.5)=0.5(75)(6.0)^2
F=0.5(75)(6.0)^2/4.5
F=300N

Answer 3

Given:
m=75kg
d=4.5m
v=6.00m/s
angle=180
Required:
F=?
Formrula:
W=change of KE
Same as:
-0.5mv^2=Fdcos()
Solve for F
F=-0.5mv^2/dcos()
F=-0.5(75kg)(6.00m/s)^2/(4.5)(cos 180)
F=-225/-4.5
F=50N

Answer 4

Wnet = 1/2mv^2 (Final) - 1/2mv^2 (Initial)
Initial = 0 because the sled started from rest, so we first need to find the final.
KE = 1/2mv^2
m = 75kg
v = 6m/s
KE = 1/2 (75)(36)^2 = 1350j
1350j is our energy, to find force we use
E = Fd so F= E/d
d = 4.5m
F= 1350/4.5 = 300N
300N is our final force, and we already know our initial force is 0
So, using Wnet = 1/2mv^2 (Final) - 1/2mv^2 (Initial) we can figure out our net force
Wnet = 300N - 0N
Wnet = 300N

Answer 5

Wnet = 1/2mv^2 (Final) - 1/2mv^2 (Initial)
Initial = 0 because the sled started from rest, so we first need to find the final.
KE = 1/2mv^2
m = 75kg
v = 6m/s
KE = 1/2 (75)(36)^2 = 1350j
1350j is our energy, to find force we use
E = Fd so F= E/d
d = 4.5m
F= 1350/4.5 = 300N
300N is our final force, and we already know our initial force is 0
So, using Wnet = 1/2mv^2 (Final) - 1/2mv^2 (Initial) we can figure out our net force
Wnet = 300N - 0N
Wnet = 300N