A small block of mass m can slide along the frictionless loop-the loop. The block is released from rest at point P, at height h =5R 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 (which has a height R) b) the top of the loop? If the gravitational poteintal 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 d) at point Q and e) at the top of the loop?
a) The block is at potental enrgy then it goes to kinetic enrgy . so would i use mgh= 1/2mv^2?
b) I don't really get what to do here.
c)When the potential energy of the block is zero at the bottom of the loop whould this really make a difference in the answers to c), d) and e)? Wouldn't it be the same equation as above?
a/ Yes, but the mgh height is from the top starting point vertically to Q.
b. PE will be mgh where h is the vertical height from the bottom to the point. For instance, at the top of the loop, h is 2R, so gpe is 2mgr
a) To find the work done by the gravitational force from point P to point Q, you can use the principle of conservation of energy. The initial potential energy at point P is given by mgh, where m is the mass of the block and h is the height of point P above the bottom of the loop. The final kinetic energy at point Q is given by 1/2mv^2, where v is the velocity of the block at point Q.
Since the loop is frictionless, the total mechanical energy (sum of potential and kinetic energies) is conserved. Therefore, the work done by the gravitational force is equal to the change in mechanical energy. Thus:
Work done = Final mechanical energy - Initial mechanical energy
Work done = (1/2)m(v^2) - mgh
b) At the top of the loop, the block reaches its maximum height and momentarily comes to rest. The work done by the gravitational force can be calculated using the same principle as in part a). The height at the top of the loop is 2R, where R is the radius of the loop. The final potential energy at the top of the loop is given by mgh, where h is 2R. The initial kinetic energy is zero since the block is momentarily at rest.
Therefore, the work done by the gravitational force at the top of the loop is:
Work done = Final mechanical energy - Initial mechanical energy
Work done = mgh - 0
c) When the gravitational potential energy of the block-Earth system is taken to be zero at the bottom of the loop, it means that the potential energy at the bottom of the loop is zero. However, this does not affect the calculation of potential energy at other points.
d) At point Q, the potential energy is given by mgh, where h is the height of point Q above the bottom of the loop, which is R.
e) At the top of the loop, the potential energy is given by mgh, where h is the height of the top of the loop above the bottom, which is 2R.
In summary:
a) Work done = (1/2)m(v^2) - mgh, where h = 5R and v is the velocity at point Q.
b) Work done = mgh, where h = 2R.
c) The zero potential energy at the bottom of the loop does not affect the calculations for points P, Q, or the top of the loop.
d) Potential energy at point Q = mgh, where h = R.
e) Potential energy at the top of the loop = mgh, where h = 2R.