## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 6 - Problems - Page 232: 87

#### Answer

Both methods lead to the same result that the tension is $27.1~N$, however, using work and energy seems to be faster and easier.

#### Work Step by Step

We can use work and energy to solve this question. Since the block is at rest at the bottom, the change in kinetic energy is zero, so the total work (from gravity and tension) must be zero. We can use work and energy to find the tension: $\sum Work = 0$ $mgd_1~sin~\theta -T~d_2 = 0$ $T~d_2 = mgd_1~sin~\theta$ $T = \frac{mgd_1~sin~\theta}{d_2}$ $T = \frac{(4.0~kg)(9.80~m/s^2)(8.0~m)~sin~15^{\circ}}{3.0~m}$ $T = 27.1~N$ We can use Newton's laws and kinematic equations to solve this question. We can find the acceleration for the first 5.0 meters: $mg~sin~\theta = ma$ $a = g~sin~\theta$ $a = (9.80~m/s^2)~sin~15^{\circ}$ $a = 2.536~m/s^2$ We can find the velocity after the block slides 5.0 meters: $v_f^2 = v_0^2+2ad$ $v_f = \sqrt{v_0^2+2ad}$ $v_f = \sqrt{0+(2)(2.536~m/s^2)(5.0~m)}$ $v_f = 5.036~m/s$ We can let this be the initial velocity for the next part of the question. We can find the rate of deceleration for the final 3.0 meters: $v_f^2 = v_0^2+2ad$ $a = \frac{v_f^2-v_0^2}{2d}$ $a = \frac{0-(5.036~m/s)^2}{(2)(3.0~m)}$ $a = -4.23~m/s^2$ We can use the magnitude of acceleration to find the tension: $T - mg~sin~\theta = ma$ $T = m~(g~sin~\theta + a)$ $T = (4.0~kg)~(9.80~m/s^2~sin~15^{\circ} + 4.23~m/s^2)$ $T = 27.1~N$ Both methods lead to the same result that the tension is $27.1~N$, however, using work and energy seems to be faster and easier.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.