Chapter 6 - Problems - Page 232: 87

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.

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.