4.6 Problem-solving strategies  (Page 3/5)

(d)

Using the free-body diagram:

$F_{\text{net}}=T-f-mg=\text{ma}$ ,

so that

$a=\frac{T-f-\text{mg}}{m}=\frac{1\text{.}\text{250}\times {\text{10}}^7\text{N}-4.50\times {\text{10}}^{\text{6}}N-(5.00\times {\text{10}}^5\text{kg})(9.{\text{80 m/s}}^2)}{5.00\times {\text{10}}^5\text{kg}}=\text{6.20}{\text{m/s}}^2$ .

1. Use Newton’s laws of motion.

   

2. Given : $a=4.00g=(4.00)(9.{\text{80 m/s}}^2)=\text{39.2}{\text{m/s}}^2\text{;}$ $m=\text{70}\text{.}\text{0 kg}$ ,

   Find: $F$ .

3. $$ $$ $\sum F\text{=+}F-w=\text{ma}\text{,}$ so that $F=\text{ma}+w=\text{ma}+\text{mg}=m(a+g)$ .

   $F=(\text{70.0 kg})[(\text{39}\text{.}{\text{2 m/s}}^2)+(9\text{.}{\text{80 m/s}}^2)]$ $=3.\text{43}\times {\text{10}}^3\text{N}$ . The force exerted by the high-jumper is actually down on the ground, but $F$ is up from the ground and makes him jump.

4. This result is reasonable, since it is quite possible for a person to exert a force of the magnitude of ${\text{10}}^3\text{N}$ .