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jeyben [28]
3 years ago
11

Charlie pulls horizontally to the right on a wagon with a force of 37.2 N. Sara pulls horizontally to the left with a force of 2

2.4 N. How much combined work is done on the wagon after it has moved 2.50 meters to the right?
Physics
2 answers:
Sidana [21]3 years ago
7 0

Answer:

The work done on the wagon is 37 joules.

Explanation:

Given that,

The force applied by Charlie to the right, F = 37.2 N

The force applied by Sara to the left, F' = 22.4 N

We need to find the work done on the wagon after it has moved 2.50 meters to the right. The net force acting on the wagon is :

F_n=F-F'

F_n=37.2-22.4

F_n=14.8\ N

Work done on the wagon is given by the product of net force and displacement. It is given by :

W=F_n\times d

W=14.8\ N\times 2.5\ m

W = 37 Joules

So, the work done on the wagon is 37 joules. Hence, this is the required solution.

Novosadov [1.4K]3 years ago
3 0

Answer:

Explanation:

Rightwards force, F1 = 37.2 N

Leftwards force, F2 = 22.4 N

distance, d = 2.5 m

Net force, F = F1 - F2

F = 37.2 - 22.4

F = 14.8 N rightwards

Work done, W = force x distance

W = 14.8 x 2.5

W = 37 J

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Answer:

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        \frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o})  (6)

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    \frac{1}{2} * 2.00 kg* v^{2}  = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)

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