Answer to Question #140545 in Mechanics | Relativity for sridhar

Question

A force of 4N acts on a 10Kg body initially at rest.Let W(1) is work done by force during 0<=t<=1s .Likewise W(2) is the work done by force during 1<=t<=2s where t is time.The ratio W(2))/W(1) is
ANS 3:1

Answer 1

$F=4(N);$

$m=10(kg);$

$t_1=1(s);\;\;t_2=2(s);$

Solution:

$a=\frac{F}{m}=\frac{4}{10}=0.4(m/s^2);$

$s=\frac{a\cdot t^2}{2};$

$W=F\cdot s=F\cdot\frac{a\cdot t^2}{2};$

$W_1=F\cdot\frac{a\cdot t_1^2}{2}=4\cdot\frac{0.4\cdot 1^2}{2};=0.8(J);$

$W_2=F\cdot\frac{a\cdot (t_2^2-t_1^2)}{2}=4\cdot\frac{0.4\cdot (2^2-1^2)}{2};=2.4(J);$

$\frac{W_2}{W_1}=\frac{2.4}{0.8}=3.$

Answer: the ratio W(2)/W(1) is 3:1.