Calculating Work Done by a Force on a Coin

In summary, a coin slides over a frictionless plane with a constant force of 8.2 N acting on it. The force is directed at an angle of 108° counterclockwise from the positive x axis. The coin travels from the origin to a point with xy coordinates (2.2 m, 6.1 m). Using the distance formula, the displacement is 6.48 m. By solving for the angle between the displacement and the positive x axis, it is determined to be 70.167°. The angle between the force and the displacement is calculated to be 38°, which is then used in the work equation to find that the work done by the force on the coin during the displacement is
  • #1
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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  • #2
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)
 

1. How do you calculate the work done by a force on a coin?

To calculate the work done by a force on a coin, you need to multiply the force applied by the distance the coin moves in the direction of the force. This can be represented by the formula W = F x d, where W is work, F is force, and d is distance.

2. What is considered a force on a coin?

A force on a coin can be any external push or pull that causes the coin to move. This can include gravitational force, frictional force, and applied force from a person or object.

3. How do you measure the distance a coin moves?

The distance a coin moves can be measured using a ruler or tape measure. Place the ruler or tape measure next to the coin and measure the distance from its starting position to its ending position.

4. Can the work done by a force on a coin be negative?

Yes, the work done by a force on a coin can be negative if the force is acting in the opposite direction of the coin's movement. This indicates that the force is doing work against the coin, rather than doing work on it.

5. What is the unit of measurement for work?

The unit of measurement for work is joules (J). This unit is equivalent to one newton-meter (N*m), where newtons represent force and meters represent distance.

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