A classical mechanics problem by Yahia El Haw

Classical Mechanics Level 2

A $1.2\text{ kg}$ mass is projected down a rough semi-circular vertical track of radius $2.0 \text{ m}$ , as shown in the diagram below. The speed of the mass at point $A$ is $3.2\text{ m/s}$ , and that at point $B$ is $6.0 \text{ m/s}$ .

How much work is done on the mass between $A$ and $B$ by the force of friction?

-8.9 J -8.1 J -7.9 J -24 J -6.6 J

1 solution

Yahia El Haw
Nov 1, 2016

Total Energy at A:

$\begin{array} { l l l } U_A + KE_A & = mgh & + \frac{1}{2} mv^2 \\ & = 1.2 \times 9.8 \times 2 & + \frac{1}{2} 1.2 \times 3.2^2 \\ & = 29.664 J \end{array}$

Total Energy at B:

$\begin{array} { l l } KE_B & = \frac{1}{2} mv^2 \\ & = \frac{1}{2} \times 1.2 \times 6^2 \\ & = 21.6 J \\ \end{array}$

Hence, the work of friction force is the change in total energy, which is

$21.6 - 29.664 = -8.1 J$

Maybe you could specify that one should talk $g=9.8 \frac{m}{s^2}$ . Only got it right because it's multiple choice and $g=10 \frac{m}{s^2}$ didn't fit any solution.

Kai Ott - 4 years, 6 months ago

Solved in the same way.

Niranjan Khanderia - 4 years, 6 months ago

A classical mechanics problem by Yahia El Haw

1 solution

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