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
