## College Physics (7th Edition)

a). $The\, diagram \,is\, as\, shown.$ b). $680.4N$ c). $2.35s$
a). The diagram is as shown. b). $F_{p}=m(g-a)$ Now, $h=0+\frac{1}{2}at^{2}$ $a=\frac{2h}{t^{2}}=\frac{2\times7}{2.5^{2}}=2.24 m/s^{2}$ Thus, $F_{p}=90(9.8-2.24)=680.4N$ c). $a=\frac{mg-F_{p}}{m}=\frac{75\times9.8-0.8\times680.4}{75}=2.54m/s^{2}$ $h=\frac{1}{2}at^{2}$ $t=\sqrt \frac{2h}{a}=\sqrt \frac{2\times7}{2.54}=2.35s$