## University Physics with Modern Physics (14th Edition)

(a) Please refer to the free body diagram below. (b) We can find the acceleration during the jump. $a = \frac{v-v_0}{t} = \frac{4.0~m/s - 0}{0.0010~s}$ $a = 4000~m/s^2$ We can use a force equation to find $F_N$. $\sum F = ma$ $F_N - mg = ma$ $F_N = ma + mg$ $F_N = (1.23\times 10^{-5}~kg)(4000~m/s^2)+(1.23\times 10^{-5}~kg)(9.80~m/s^2)$ $F_N = 0.0493~N$ The ground exerts a force of 0.0493 N on the froghopper. (c) We can express the force in terms of the froghopper's weight $mg$. $F = \frac{0.0493~N}{(1.23\times 10^{-5}~kg)(9.80~m/s^2)}$ $F = 409\times ~weight$ The normal force is 409 times the froghopper's weight.