Math Problem Statement

In the figure, a small block of mass m = 0.015 kg can slide along the frictionless loop-the-loop with radius R = 19 cm. The block is released from rest at a height h = 5R above the bottom of the loop. How much work does the gravitational force do as the block travels from point P to (a) point Q and (b) the top of the loop? Calculate potential energy at point P, Q, and the top of the loop assuming gravitational potential energy is zero at the bottom of the loop.

Solution

The problem involves a small block of mass m=0.015kgm = 0.015 \, \text{kg} sliding along a frictionless loop-the-loop with a radius R=19cm=0.19mR = 19 \, \text{cm} = 0.19 \, \text{m}. The block is released from rest at a height h=5R=5×0.19m=0.95mh = 5R = 5 \times 0.19 \, \text{m} = 0.95 \, \text{m} above the bottom of the loop.

We need to calculate the following:

  1. The gravitational potential energy at different points (a, b, c, d, e).
  2. The work done by gravity as the block moves from one point to another.

Let's break down the required calculations:

Gravitational Potential Energy (GPE)

The GPE at any point is given by the formula: GPE=mgh\text{GPE} = mgh where mm is the mass of the block, g=9.8m/s2g = 9.8 \, \text{m/s}^2 is the acceleration due to gravity, and hh is the height relative to the bottom of the loop.

Points in the Problem

  • Point P: The block is released at a height h=5R=0.95mh = 5R = 0.95 \, \text{m}.
  • Point Q: This is at the same height as the center of the loop, so its height is h=2R=0.38mh = 2R = 0.38 \, \text{m}.
  • Top of the loop: The height here is h=2R=0.38mh = 2R = 0.38 \, \text{m}.

Gravitational Potential Energy at each point:

  • At point P: GPEP=mghP=(0.015)(9.8)(0.95)=0.13965J\text{GPE}_P = mgh_P = (0.015)(9.8)(0.95) = 0.13965 \, \text{J}

  • At point Q: GPEQ=mghQ=(0.015)(9.8)(0.38)=0.05586J\text{GPE}_Q = mgh_Q = (0.015)(9.8)(0.38) = 0.05586 \, \text{J}

  • At the top of the loop: GPEtop=mghtop=(0.015)(9.8)(0.38)=0.05586J\text{GPE}_{\text{top}} = mgh_{\text{top}} = (0.015)(9.8)(0.38) = 0.05586 \, \text{J}

Work done by gravity from point P to Q:

The work done by gravity is the change in potential energy: W=GPEPGPEQ=0.13965J0.05586J=0.08379JW = \text{GPE}_P - \text{GPE}_Q = 0.13965 \, \text{J} - 0.05586 \, \text{J} = 0.08379 \, \text{J}

Work done by gravity from point P to the top of the loop:

W=GPEPGPEtop=0.13965J0.05586J=0.08379JW = \text{GPE}_P - \text{GPE}_{\text{top}} = 0.13965 \, \text{J} - 0.05586 \, \text{J} = 0.08379 \, \text{J}

Now, matching this with your input:

  • (a) and (b) seem to ask for the work done from point P to point Q and P to the top of the loop, and they should both be positive values 0.08379J0.08379 \, \text{J} instead of negative.
  • (c) seems correct at 0.13965J0.13965 \, \text{J}, which is the GPE at point P.
  • (d) and (e) are both 0.05586J0.05586 \, \text{J}, which is correct for both point Q and the top of the loop.

Let me know if you need more details or have questions!

Five Related Questions:

  1. How would the work done change if friction were involved in the loop?
  2. What is the speed of the block at the bottom of the loop?
  3. How would the potential energy at the top change if the mass of the block were doubled?
  4. What happens to the work done if the block were released from a height of 3R3R?
  5. Can we calculate the normal force acting on the block at the top of the loop?

Tip:

Always keep track of height changes when calculating potential energy, as it directly influences the work done by gravity!

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Math Problem Analysis

Mathematical Concepts

Gravitational potential energy
Work-energy theorem
Conservation of mechanical energy

Formulas

Gravitational potential energy (GPE) = mgh
Work done by gravity = Change in gravitational potential energy

Theorems

Work-Energy Theorem: The work done by forces on an object equals the change in its mechanical energy.
Conservation of Mechanical Energy: In a frictionless system, the total mechanical energy (kinetic + potential) is conserved.

Suitable Grade Level

Undergraduate Physics