Help with a Potential Energy Problem

[champ0342]

In the figure below, a small block of mass m = 1.0 kg can slide along the frictionless loop-the-loop. The block is released from rest at point P, at height h = 8R above the bottom of the loop. (For all parts, answer using g for the acceleration due to gravity, and R and m as appropriate.)
"[URL
http://www.webassign.net/hrw/08_33.gif

(a) How much work does the weight of the block do on the block as the block travels from point P to point Q?

(b) How much work does the weight of the block do on the block as the block travels from point P to the top of the loop?

(c) What is the potential energy when the block is at point P? (Assume that the gravitational potential energy of the block Earth system is taken to be zero at the bottom of the loop.)

(d) What is the potential energy when the block is at point Q?

(e) What is the potential energy when the block is at the top of the loop?

 

[Google_Spider]

Did you read https://www.physicsforums.com/showthread.php?t=94379" ??

 

[Moetasim]

(a) The work done by the weight of the block on the block as it travels from point P to point Q can be calculated using the equation W = mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the change in height. In this case, the change in height is R, since the block is traveling from a height of 8R to a height of 9R. Therefore, the work done by the weight of the block is W = (1.0 kg)(9.8 m/s^2)(R) = 9.8R Joules.

(b) As the block travels from point P to the top of the loop, the work done by the weight of the block is the same as in part (a), since the height change is still R. Therefore, the work done by the weight of the block is again 9.8R Joules.

(c) The potential energy of the block at point P can be calculated using the equation PE = mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height of the block. In this case, the height of the block is 8R, so the potential energy at point P is PE = (1.0 kg)(9.8 m/s^2)(8R) = 78.4R Joules.

(d) At point Q, the height of the block is 9R, so the potential energy can be calculated using the same equation as in part (c). Therefore, the potential energy at point Q is PE = (1.0 kg)(9.8 m/s^2)(9R) = 88.2R Joules.

(e) At the top of the loop, the height of the block is 10R. Therefore, the potential energy can be calculated using the same equation as in parts (c) and (d). The potential energy at the top of the loop is PE = (1.0 kg)(9.8 m/s^2)(10R) = 98R Joules.