h = (8 meters)*sin(12

^{o}) = 1.66 m

^{o}) = 44.95N

^{2})*(1.66 m)

A ramp is inclined at 12 degrees to the horizontal. A cart is pulled 8m up the ramp by a force of 120N, at an angle of 10 degrees to the surface of the ramp. Determine the mechanical work done against gravity. Do not use physics formulas to solve this... use calculus and vectors.

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The cart is lifted to a height, h:

h = (8 meters)*sin(12^{o}) = 1.66 m

h = (8 meters)*sin(12

The force in the vertical direction is:

Fy = 120N*sin(22^{o}) = 44.95N

Work = F*d = (44.95)*(1.66) = 74.62 joules

----------

74.62 J = Mgh

Where M is the cart's mass, g is the acceleration due to gravity, and h is the height the cart is pulled up.

74.62 J= M*(9.8 m/s^{2})*(1.66 m)

4.59 kg = M, the mass of the cart

The angle of pull is 12º + 10º above the horizontal so **F**_{y} = 120N•sin(22º)

is the component of the force in the vertical direction pulling against gravity.

Mechanical work is only done on an object when the force and motion are parallel.

Therefore the work is **F**_{y} times the distance in the vertical direction.

The cart moved 8 m (I assume 8 m is the distance parallel to the ramp) at 12º.

So the vertical distance the cart moves is **X**_{y}

sin 12º = **X**_{y} / 8 m ; **X**_{y} = 8m•sin(12º)

Therefore work = **F**_{y} • **X**_{y} = 120N•sin(22º)•8m•sin(12º) = 960•sin(22º)•sin(12º) N·m

You can't calculate work without at least 1 physics formula that W = F·d where f & d are parallel.

All the bold quantities are vector quantities.

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