## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

(a) $F = 225~N$ (b) $v = 0.200~m/s$
(a) We can find the acceleration of the rock. $a = \frac{v^2-v_0^2}{2d}$ $a = \frac{(30~m/s)^2-0}{(2)(1.0~m)}$ $a = 450~m/s^2$ We can use the acceleration to find the force that Bob exerts on the rock. $F = ma$ $F = (0.500~kg)(450~m/s^2)$ $F = 225~N$ (b) By Newton's third law, the rock exerts an equal and opposite force on Bob. We can find Bob's acceleration. $F = ma$ $a = \frac{F}{m}$ $a = \frac{225~N}{75~kg}$ $a = 3.0~m/s^2$ We can find the time it takes for the rock to move through a distance of 1.0 meter while Bob is throwing it. $d = \frac{1}{2}at^2$ $t = \sqrt{\frac{2d}{a}}$ $t = \sqrt{\frac{(2)(1.0~m)}{450~m/s^2}}$ $t = 0.0667~s$ We can find Bob's speed after releasing the rock. $v = v_0+at$ $v = 0 + (3.0~m/s^2)(0.0667~s)$ $v = 0.200~m/s$