Calculating Work Done by a Force on a Coin

AI Thread Summary
A coin moves across a frictionless plane from the origin to the coordinates (2.2 m, 6.1 m) under the influence of an 8.2 N force at a 108° angle from the x-axis. The displacement is calculated using the distance formula, resulting in approximately 6.48 m. The angle between the force and displacement is determined to be 38° after finding the angle of displacement as 70.167°. The work done by the force is then calculated using the formula W = FdCos(a), yielding a result of 42 Joules. This confirms the correct approach to solving the problem.
dorkymichelle
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Homework Statement


A coin slides over a frictionless plane and across an xy coordinate system from the origin to a point with xy coordinates (2.2 m, 6.1 m) while a constant force acts on it. The force has magnitude 8.2 N and is directed at a counterclockwise angle of 108° from the positive direction of the x axis. How much work is done by the force on the coin during the displacement?

Homework Equations


W=FdCos a
a = angle between force and displacement
x2+y2 = d2

The Attempt at a Solution



pt 1 = 0,0, pt 2 = 2.2m, 6.1m
using distance formula
displacement = square root (2.2)2+6.12
d= 6.48
force = 8.2
and a = 108 degrees
so w = 8.2*6.48 *cos 108 = -16.42 J
I think the problem is that the angle between displacement and force is not really 108 and I am supposed to figure out where it is.. not I'm not sure where that's going..

edit , I just did tan b = 6.1/2.2
b = 70.167 degrees, being the angle between the displacement and pos x axis.
would the angle a, between force and displacement be angle of force - angle of displacement? 108-70 = 38 degrees.
and 38 degrees would be the angle that is put into the w = f*d equation?
I need someone to confirm because I'm on my last try on homework = /
 
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yup. looks good. I'll do it quick here to make sure we both get the same answer. One sec..

I get 42 Joules (to 2 sig figs)
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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