The issue is with the method that you used. Instead of finding the magnitude of the force, you need to find the component of the force that is in the direction of the displacement.
The displacement vector is given by:
d = 50(cos(20) i + sin(20) j)
Now, we need to find the component of the force that is in the direction of the displacement. This is done using the dot product.
Work done, W = F . d
W = (5 i + 6 j + 1 k) . (50cos(20) i + 50sin(20) j)
W = 5 * 50cos(20) + 6 * 50sin(20)
W = 250(5cos(20) + 6sin(20))
Now, just calculate this value:
W = 250(5 * 0.9397 + 6 * 0.34202)
W = 250(4.6985 + 2.05212)
W = 250(6.75062)
W = 1687.655
However, the answer you're looking for is in Joules, so we need to divide by 5 (1 J = 5 Nm):
W = 1687.655 / 5
W ≈ 337.53 J
As we can see, the answer is approximately 337.5 J.
The force Fa=(5i+6j+1k) is applied on a material point while it is shifted by 50m in Oxy plane along a direction 20o CCW versus Ox axis. Find the work done by force on the particle during this shift.
The answer should be 337.5J
What I did:
I found the magnitude of the force: 5^2+6^2+1^=62
sqrt(62)=7.874
I then multiplied the answer by 50(the shift) and then multiplied it by cos20 to find the work done, but I got 369.95J, what did I do wrong??
1 answer
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