Newtons Laws » Force on a Small problem?
A physics class pushes a small wagon along a flat horizontal parking lot with a force of 75.02 N. Starting from rest the wagon travels 12.4 meters in 13.5 seconds. What is the wagon's acceleration?
What is the magnitude of the force of friction acting on the wagon in the previous question, if the loaded wagon has a mass of 200.0 kg?
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Verified answer
1)By s = ut + 1/2at^2
=>12.4 = 0 + 1/2 x a x (13.5)^2
=>a = 0.14 m/s^2
2) F(net) = m x a = 200 x 0.14 = 27.22 Newton
=>By F(net) = F(applied) - F(friction)
=>F(friction) = F(applied) - F(net)
=>F(friction) = 75.02 - 27.22 = 47.80 Newton
distance = 1/2 acceleration time^2
Solving for a we find
2d/(t^2) = a so a = .136m/s^2
Force = Mass x Acceleration
Since we know the Acceleration and Mass we can figure out the Force without the friction
200kg x .136 m/s^2
27.22 N
So the force of Friction = Total Force - the force to we just calculated
47.8 Newtons
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