Answer:
1.29 m
Explanation:
Net force acting on the dog, [tex]F_net[/tex] will be given by
[tex]F_net=F_upward – F_downward=2.1F_g –F_g=2.1mg-mg=1.1(mg) [/tex] 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
[tex]\triangle W=\triangle PE[/tex]
[tex]F_net \triangle x=mg(\triangle y)[/tex]
[tex]1.1mg(x-\hat x)= mg(\triangle y)[/tex]
[tex]1.1(x-\hat x)=\triangle y[/tex]
[tex]1.1(0.5 m-0.1 m)=\triangle y[/tex]
[tex]1.1(0.4 m) =\triangle y[/tex]
[tex]\triangle y =0.44 m[/tex]
Height which the dog can jump, [tex]h=x+\triangle y[/tex]
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
He can jump "1.29 m" high.
According to the question,
The net force on dog be:
→ [tex]F_{net} = F_{upward} \ \hat a \ F_{downward}[/tex]
[tex]= 2.1 F_g \ \hat a \ F_g[/tex]
[tex]= 2.1 \ mg- mg[/tex]
[tex]= 1.1(mg)[/tex]
We know the relation,
→ [tex]\Delta W = \Delta PE[/tex]
[tex]F_{net} \Delta x= mg (\Delta y)[/tex]
By substituting the values,
[tex]1.1 mg(x-\hat x) = mg(\Delta y)[/tex]
[tex]1.1 (0.5 \ m -0.1 \ m) = \Delta y[/tex]
[tex]1.1\times 0.4 m= \Delta y[/tex]
[tex]\Delta y = 0.44 \ m[/tex]
Now,
The height which the dog can run will be:
→ [tex]h = x+ \Delta y[/tex]
[tex]= 0.6+0.44[/tex]
[tex]= 1.04 \ m[/tex]
hence,
The total height be:
= Height to jump + Half height of the dog
= [tex]1.04+0.5(0.5)[/tex]
= [tex]1.29 \ m[/tex]
Thus the above answer is correct.
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