Question

An athlete is running at a constant velocity with a javelin held in his right hand. The force he is applying on the javelin as he carries it is 5.0 newtons. If he covers a distance of 10 meters, what work has he done on the javelin?

Answer 1

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Since the force he is applying is 5N and he carries it a distance of 10 meters and the displacement and the force are assumed to be in the same direction we have 5*10 so 50 J

Answer 2

Answer:

He has done a work of 50 J on the javelin.

Explanation:

The work can be defined as:

$W = F \cdot d \cos \theta$ (1)

Where W is the work, d is the distance and $\theta$ is the angle between the force and the direction of the displacement.

For this particular case the force has the same direction of the displacement, so $\theta = 0^{\circ}$

Then, replacing all the values in equation 1 it is gotten:

$W = (5.0 N) \cdot (10 m) \cos 0^{\circ}$

$W = 50 N.m$

$W = 50 Kg.m/s^{2}.m$

$W = 50 Kg.m^{2}/s^{2}$

But 1J = $Kg.m^{2}/s^{2}$, therefore:

$W = 50 J$

Hence, he has done a work of 50 J on the javelin.

Key term:

1J = 1 Joule