Asmall block of mass m = 0.032 kg can slide along the frictionless loop-the-loop, with loop radius r = 12 cm. the block is released from rest at point p, at height h = 5.0r above the bottom of the loop. how much work does the gravitational force do on the block as the block travels from point p to:
a) point q? (0.15 j)
b) the top of the loop? (0.11 j) if the gravitational potential energy of the block-earth system is taken to be zero at the bottom of the loop, what is that potential energy when the block is:
c) at point p? (0.19 j) d) at point q? (0.038 j) e) at the top of the loop? (0.075 j) f) if, instead of merely being released, the block is given some initial speed downward along the track, do the answers to a) through c) increase, decrease or remain the same? (same)

Change in gravitational potential is used to move the block from a high

elevation to a lower elevation.

a) Work done travelling from P to Q is approximately 0.15 J

b) Work done travelling from P to the top of the loop is approximately 0.11 J

c) Potential energy at point P is 0.19 J

d) Potential energy at point Q is 0.038 J

e) Potential energy at the top of the loop, is 0.075 J

f) The work done by gravity is the same, therefore, the values are the same

Reasons:

The given parameters are;

Mass of the block, m = 0.032 kg

Radius of loop, R = 12 cm = 0.12 m

Height of point P = 5.0·R

a) Work done by gravity force by moving from P to point Q·

The height of the point Q = R

∴ Change in height, Δh = 5.0·R - R = 4.0·R

∴ Δh = 4.0 × 0.12 m = 0.48 m

The work done, W = m·g·Δh

∴ W = 0.032 kg × 9.81 m/s² × 0.48 m ≈ 0.15 J

Work done by gravity force by moving from P to point Q, W = 0.15 J

b) The elevation of the top of the loop, = 2·R

Change in elevation, Δh = 5.0·R - 2.0·R = 3.0·R

Work done W = m·g·Δh

∴ Work done W = 0.032 × 9.81 × 3.0 × 0.12 ≈ 0.11

Work done by travelling from P to the top of the loop ≈ 0.11 J

c) The potential energy, P.E. = m·g·h

∴ P.E. 0.032 kg × 9.81 m/s² × 5 × 0.12 m≈ 0.19 J

The potential energy at point P ≈ 0.19 J

d) At point Q, we have;

The height, h = R = 0.12 m

P.E. = 0.032 kg × 9.81 m/s² × 0.12 m ≈ 0.038 J

e) At the top of the loop, h = 2.0·R = 2.0 × 0.12 m = 0.24 m

P.E. = 0.032 kg × 9.81 m/s² × 0.24 m ≈ 0.075 J

f) The work done by gravity is given by the change in elevation, resulting

in change in potential energy, therefore, given that the change in elevation

and therefore, potential energy in part a) through c) are the same, the work

done by gravity is therefore, the same in each case.

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Part a)

Part b)

Part c)

Part d)

Part e)

Part f)

It is independent of the speed of the object so all part answers will remain the same

Explanation:

Part a)

As we know that Point P is at height 5R while point Q is at height R

so the work done by gravity from P to Q is given as

Part b)

When it reaches to the top of the loop then its final height from ground is

h = 2R

so work done from P to Q is given as

Part c)

Potential energy at P point is given as

Part d)

Potential energy at Q point is given as

Part e)

Potential energy at top point is given as

Part f)

Since all the answer from part a) to part e) depends only upon the position of the object.

So here we can say that it is independent of the speed of the object so all part answers will remain the same

i would say the answer is food.

hope this !