Questions: In the figure, a small block of mass m=0.027 kg can slide along the frictionless loop-the-loop, with loop radius R=18 cm. The block is released from rest at point P, at height h=3 R 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 and (b) the top of the loop? 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, (d) at point Q, and (e) at the top of the loop? (a) Number Units (b) Number Units (c) Number Units (d) Number Units (e) Number Units

In the figure, a small block of mass m=0.027 kg can slide along the frictionless loop-the-loop, with loop radius R=18 cm. The block is released from rest at point P, at height h=3 R 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 and (b) the top of the loop? 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, (d) at point Q, and (e) at the top of the loop?
(a) Number Units
(b) Number Units
(c) Number Units
(d) Number Units
(e) Number Units

Transcript text: In the figure, a small block of mass $m=0.027 \mathrm{~kg}$ can slide along the frictionless loop-the-loop, with loop radius $R=18 \mathrm{~cm}$. The block is released from rest at point $P$, at height $h=3 R$ 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 $Q$ and (b) the top of the loop? 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, (d) at point Q. and (e) at the top of the loop? (a) Number i Units $\square$ (b) Number $\square$ Units $\square$ (c) Number $\square$ Units $\square$ (d) Number $\square$ Units $\square$ (e) Number $\square$ Units $\square$

Solution

Solution Steps

Step 1: Calculate the work done by gravity from P to Q.

The work done by gravity is given by the change in gravitational potential energy. Since the block moves vertically downwards by a distance of _h_ - _R_ = 3_R_ - _R_ = 2_R_ when going from P to Q, the work done by gravity is _mgh_ where _h_ = 2_R_. Therefore, W_PQ = _mg_(2_R_) = 2_mgR_. Substituting _m_ = 0.027 kg, _g_ = 9.8 m/s², and _R_ = 0.18 m, we get: W_PQ = 2 * 0.027 kg * 9.8 m/s² * 0.18 m = 0.095256 J.

Step 2: Calculate the work done by gravity from P to the top of the loop.

The vertical displacement from P to the top of the loop is _h_ - 2_R_ = 3_R_ - 2_R_ = _R_. Therefore, the work done by gravity is W_PTop = _mgR_. Substituting the values, we get: W_PTop = 0.027 kg * 9.8 m/s² * 0.18 m = 0.047628 J.

Step 3: Calculate the potential energy at P, Q, and the top of the loop.

The potential energy is given by _mgh_, where _h_ is the height above the bottom of the loop. (c) At point P, _h_ = 3_R_, so U_P = 3_mgR_ = 3 * 0.047628 J = 0.142884 J. (d) At point Q, _h_ = _R_, so U_Q = _mgR_ = 0.047628 J. (e) At the top of the loop, _h_ = 2_R_, so U_Top = 2_mgR_ = 2 * 0.047628 J = 0.095256 J.

Final Answer

(a) 0.095 J (b) 0.048 J (c) 0.14 J (d) 0.048 J (e) 0.095 J

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