A little red wagon with mass 7.00 kg moves in a straight line on a frictionless horizontal surface. It has an initial speed of 4.00 m/s and then is pushed 3.0 m in the direction of the initial velocity by a force with a magnitude of 10.0 N.
a) Use the work-energy theorem to calculate the wagon’s final speed.
b) Calculate the acceleration produced by the force. Use this acceleration to calculate the wagon’s final speed. Compare this result to that calculated in part (a)
(a) The work done by the applied force over the 3.00m distance is:
W = FΔx
= (10.0N)(3.00m)
= 30.0J
The work-energy theorem says that the net work done on an object is equal to the change in that object’s kinetic energy. Since the problem gives no other forces other than the 10.0N force acting on the wagon, then the net work done is 30.0J. The final velocity can now be found:
ΣW = 0.5mv² – 0.5mv₀²
v = √[(2ΣW + mv₀²) / m]
= √[{2(30.0J) + (7.00kg)(4.00m/s)²} / 7.00kg]
= 4.96m/s
(b) Newton’s 2nd law gives the acceleration of the wagon as:
ΣF = ma
a = ΣF / m
= 10.0N / 7.00kg
= 1.43m/s²
The wagon’s final speed by kinematics is:
v² = v₀² + 2aΔx
v = √[(4.00m/s)² + 2(1.43m/s²)(3.00m)]
= 4.96m/s
Exactly the same! Hope this helps.
(a) The work done by the applied force over the 3.00m distance is:
W = FΔx
= (10.0N)(3.00m)
= 30.0J
The work-energy theorem says that the net work done on an object is equal to the change in that object’s kinetic energy. Since the problem gives no other forces other than the 10.0N force acting on the wagon, then the net work done is 30.0J. The final velocity can now be found:
ΣW = 0.5mv² – 0.5mv₀²
v = √[(2ΣW + mv₀²) / m]
= √[{2(30.0J) + (7.00kg)(4.00m/s)²} / 7.00kg]
= 4.96m/s
(b) Newton’s 2nd law gives the acceleration of the wagon as:
ΣF = ma
a = ΣF / m
= 10.0N / 7.00kg
= 1.43m/s²
The wagon’s final speed by kinematics is:
v² = v₀² + 2aΔx
v = √[(4.00m/s)² + 2(1.43m/s²)(3.00m)]
= 4.96m/s
Exactly the same! Hope this helps.
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