Chapter 6 - Applications of Integration - 6.7 Physical Applications - 6.7 Exercises - Page 470: 56

a) $1600 \ J$ b) $1600 \ J$

We are given that $x(t)=4t^2$ and $m= 2 \ kg$ and $x''(t)=8$ a) $F=ma=mx''(t)=16$ Now, $W=\int_0^{100} 16 dx =1600 \ J$ b) We are given that $x(t)=4t^2$ and $m= 2 \ kg$ and $x''(t)=8$ Now, $dx =(4t^2)^{'} =8t dt$ So,$W=\int_a^b F(x) dx= \int_0^5 (16)(8t) dt \\=64|t^2|_0^5\\=1600 \ J$