# Can someone help me with physics that has to do with work [Solved] Between points A and B, how much work is done on the rock by the normal force
-There is no work done by the normal force because all of the work is being done by gravity.
W=0J

Between points A and B, how much work is done on the rock by gravity
-Work done by gravity is just the gravitational potential energy lost by moving in the vertical direction, so you use the equation Wg=mgh. In this case, h is just the radius so Ill sub that instead
Wg=mgR
Wg=(.26)*(9.8)*(.50)
Wg=1.274 J

What is the speed of the rock as it reaches point B
-Find the speed by using the equation W=(.5)mv^2. For this part, you have to take into account the friction, Wt=Wg-Wf. So, solving for v:
v=sqrt((Wg-Wf)/(.5*m))
v=sqrt((1.274-.22)/(.5*.26))
v=2.85

Just as the rock reaches point B, what is the normal force on it due to the bottom of the bowl
-When any object reaches the bottom of a loop or circle, the normal force is equal to the weight(mg) added to the mass times the radial acceleration(Arad). The equation looks like this: n=mg m(Arad). First, you need to find the radial acceleration which is equal to v^2/R.

Now, you just plug in everything you know:
n=(.26*9.8) (.26*16.25)
n=6.77N

Equals the weight of the rock = 0.26*9.8 = 2.6 N
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Equals the weight of the rock = 0.26*9.8 = 2.6 N
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Now, you just plug in everything you know:
n=(.26*9.8) (.26*16.25)
n=6.77N

Work done by gravity is force * displacement = mgR =0.26*9.8*0.5 = 1.274 J

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0.5mv^2 = mgR work done against friction
0.5*0.26 v^2 = 1.274 0.22 =1.054 J

v = 2.85 m/s

0.5mv^2 = mgR work done against friction
0.5*0.26 v^2 = 1.274 0.22 =1.054 J

Equals the weight of the rock = 0.26*9.8 = 2.6 N
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This post is last updated on hrtanswers.com at Date : 1st of September – 2022