wagon withmass 7kg moves in a straight line on horizontal surface it has an initial speed of 4m/s and pushed 3m in the direction of initial velocity by a force 10N calculate final speed
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2 answers · Physics
Best Answer
(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.
W = FΔx
= (10.0N)(3.00m)
= 30.0J
F= Force
∆x= distance
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Work Energy Theorem
ΣW = 0.5mv² - 0.5mv₀² |
___________________
m= mass v = initial speed v=final speed
ΣW = 0.5mv² - 0.5mv₀² |
ΣW+0.5mv₀² = 0.5mv²
v²= (ΣW+0.5mv₀²)/.5m
v = √[(2ΣW + mv₀²) / m]
= √[{2(30.0J) + (7.00kg)(4.00m/s)²} / 7.00kg]
= 4.96m/s
_______________________
F