A small rock with mass 0.22 kg is released from rest at point A, which is at the

Question

A small rock with mass 0.22 kg is released from rest at point A, which is at the top edge of a large, hemispherical bowl with radius R = 0.48 m (see figure). Assume that the size of the rock is small compared to R, so that the rock can be treated as a particle, and assume that the rock slides rather than rolls. The work done by friction on the rock when it moves from points to point B at the bottom of the bowl has magnitude 0.19 J. (a) Between points A and B, how much work is done on the rock by the following. (i) the normal force J (ii) gravity J (b) What is the speed of the rock as it reaches point B? m/s (d) Just as the rock reaches point B, what is the normal force on it due to the bottom of the bowl? N

Explanation / Answer

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 I'll sub that instead
Wg=mgR
Wg=(.22)*(9.8)*(.48)
Wg=1.035 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((1.035-.19)/(.5*.22))
v=2.77m/s


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.
Arad=v^2/R
Arad=(2.77^2)/.48
Arad= 15.98

Now, you just plug in everything you know:
n=mg+m(Arad)
n=(.22*9.8)+(.22*15.98)
n=5.6716N

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