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Vladimir79 [104]
3 years ago
9

Force F1 acts on a particle and does work W1. Force F2 acts simultaneously on the particle and does work W2. The speed of the pa

rticle does not change. Which one of the following must be true?a. W1 is zero, and W2 is positiveb. W1 = - W2c. W1 is positive, and W2 is positived. W1 is positive, and W2 is zero
Physics
1 answer:
egoroff_w [7]3 years ago
7 0

Answer:

b.) W1 = -W2

Explanation:

According to Newton's third law of motion, action and reaction are equal and opposite. For the particle to maintain a constant speed, it means that the two workdone on the particle are of eqaul magnitude but act in opposite direction. The two weights, instead of adding up annul each other and has no effect on the speed of the particle.

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r(t) = (t -sin(t)i + (1 -cos(t)j,\\\\\frac{dr}{dt}=(1-cos t)i+sin t\  j\\\\dr=(1-cos\  t)i +sin\  t \ j)dt\\\\F(x,y)=x\ i +(y+6)j\\\\F(r(t))=(t-sin \ t) i+((1-cos \t)+2)j\\\\\\

F(r(t))=(t-sin \ t)i+(3-cos \ t)j\\\\\\W=\int\limits_C F.dr\\=\int\limits^{6\pi}_0(t-sin \ t)i+(3-cos \ t)j).((1-cos \ t)i+sin \ t \ j)dt\\\\=\int\limits^{6\pi}_0(t-sin \ t)(1-cos \ t)+(3-cos \ t)sin \ t \ dt\\\\=\int\limits^{6\pi}_0t-tcos \ t+2sin\ t\ dt\\\\=\int\limits^{6\pi}_0-tcos \ t\ dt+[\frac{t^2}{2}-2cos \ t]\limits^{6\pi}_0\\\\=-I+177.65

#Integrating I by parts:

I=\int\limits^{6\pi}_0 tcos \ t \ dt\\\\=[\int \ tcos \ t \ dt]\limits^{6\pi}_0\\\\=[t\int cos \ t \ dt-\int (\frac{dt}{dt}\intcos \ t \ dt)dt]\limits^{6\pi}_0\\\\=[tsin\ t -\int sin\ t \ dt]\limits^{6\pi}_0\\\\=0

W=0+177.65

Hence, work done is 177.65

3 0
3 years ago
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