Answer
(a) Please refer to the free body diagram below.
(b) The ground exerts a force of 0.0493 N on the froghopper.
(c) The normal force is 409 times the froghopper's weight.
Work Step by Step
(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.
