A 9.00 -kg particle moves from the origin to position C , having coordinates x =

Question

A 9.00-kg particle moves from the origin to position C, having coordinates x = 5.50 m and y = 5.50 m as shown in the figure. One force on the particle is the gravitational force acting in the negative y direction.

A 9.00-kg particle moves from the origin to position C, having coordinates x = 5.50 m and y = 5.50 m as shown in the figure. One force on the particle is the gravitational force acting in the negative y direction. Using the equation (W = F?r cos ? = Farrowbold·?rarrowbold), calculate the work done by the gravitational force in going from O to C along the following paths. the purple path (OAC) J the red path (OBC) the blue path (OC) J

Explanation / Answer

Realize that the work done by gravity is alligned vertically along the y-axis. In other words, we don't care what the horizontal, x, component is as any work done in the x direction will not be done by gravity.

Then realize that work is equal to force * distance, W=f*d.

What is the force of gravity acting on the particle?

The particle's mass is 9.00kg. g= accelleration = -9.8m/s^2.

force = mass * acelleration

f = 9.00kg * -9.8m/s^2

f = -88.2 N

We have the force. Now we need the distance traveled in the y direction.

It starts at the origin, y=0. It ends at 5.5m.

W = f * d

W = -88.2N * 5.5m

W = 485.1 J

Regardless of which path you take, the change in the y-component/height will be the same so the work done is independent of the path. i.e. a, b, and c are all 485.1J

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