Question

A 14-kg dog jumps up in the air to catch a ball. The dog's center of mass is normally 0.20 m above the ground, and he is 0.50 m long. The lowest he can get his center of mass is 0.10 m above the ground, and the highest he can get it before he can no longer push against the ground is 0.60 m .If the maximum force the dog can exert on the ground in pushing off is 2.1 times the gravitational force Earth exerts on him, how high can he jump?

Answer 1

Answer:

1.29 m

Explanation:

Net force acting on the dog, $F_net$ will be given by

$F_net=F_upward – F_downward=2.1F_g –F_g=2.1mg-mg=1.1(mg)$ where m is mass and g is acceleration due to gravity

Since the work done by the net force against the ground equals the final potential energy

$\triangle W=\triangle PE$

$F_net \triangle x=mg(\triangle y)$

$1.1mg(x-\hat x)= mg(\triangle y)$

$1.1(x-\hat x)=\triangle y$

$1.1(0.5 m-0.1 m)=\triangle y$

$1.1(0.4 m) =\triangle y$

$\triangle y =0.44 m$

Height which the dog can jump, $h=x+\triangle y$

H=0.6+0.44=1.04 m

Total height to reach=Height to jump+ half height of the dog

=1.04 m+0.5(0.5 m)=1.29 m