QUESTION: Dear Sir,
U am a student of tenth grade. We have a numerical problem in our book which I fail to solve. It is this:
"A man runs 50m on a horizontal road and then 5m on an inclined road with inclination 3 in 4. The weight of the man is 60kgf.Calculate the total work done by him"
First of all let me tell you what I've understood. The first part of question is simple. The work done by him when he runs on a horizontal road is 600N ×50m = 30000J. But the inclination part confuses me. Please help me out.
ANSWER: Hello AB,
The Physics definition of work is force*displacement*cosine(the angle between those 2 vectors). Since cos90=0, he gets no credit for the horizontal part of the run. You could carry 100 N weight 100 m horizontally at constant speed, and your arms might be quite tired, but that does not count as work. Energy must be conserved.
He does get credit for doing work in the inclined part of his run. The Joules of work done is represented in his increase in gravitational potential energy (which is m*g*h). The angle that is used in the cosine function in the work formula in this case is
arctan(opposite/adjacent) or arctan(3/4). So do the arctan (also called tan^-1) to get the angel, then plug the 500 N, the 5 m, and the angle into the work formula. Another way to think of this: the displacement*cos part of the formula calculates the vertical component of that uphill distance.
To verify your answer, m*g*h should give you the same answer.
I hope this helps,
---------- FOLLOW-UP ----------
QUESTION: Sir your answer really helped. But I have another doubt using arctan(3/4) I found the value of the angle. Then substituting this angle in the formula of W=F*S*cosine of angle I found the result as 2400J. However the value of m*g*h gives 1800J.
I should have drawn a picture. I tried to picture it in my head - not trustworthy. It should have been arctan(4/3).
The 2 vectors are the 600 N force pointed vertically up and the 5 m displacement along the incline. To draw those vectors (as I should have done), start both from the same point. This simple sketch is an approximation:
The arrowheads would be at the top end of the vertical line and the top-right end of the diagonal line. The angle we need is between them and is arctan(4/3). Now both calculations will give you 1800 J.
Pardon my carelessness,