Respuesta :
Answer:
2.24 m/s
Explanation:
resolving force of 29.2 N in x component
Fx = 29.2 cos 57.7
Fx = 15.6N
as force of friction is 12.7 N hence net force which produces acceleration is
15.6-12.7=2.9 N
by Newton 's law a=f/m
a= 2.9/6.87=0.422 m/s^2
now equation of motion is
v^2= U^2+2as
= 0^2+2(.422)(5.93)
v^2=5.00
v=2.24 m/s
The final speed of the suitcase is; 2.24 m.s
What is the final speed?
we are told that the suitcase is initially at rest with a force of 29.2 N at an angle of 57.7º. Thus, when we resolve along the x-direction we have;
F_x = 29.2 cos 57.7
F_x = 15.6N
Since friction of 12.7N pulss back the 15.6 N force, then;
Net force in the x-direction is; 15.6 - 12.7 = 2.9 N
Since mass is 6.87 kg, then;
Acceleration; a = F/m = 2.9/6.87 = 0.422 m/s²
From newton's second equation of motion we know that;
v² = u² + 2as
we are given;
distance; s = 5.93 m
initial speed; u = 0 m/s
Thus;
v = 0² + 2(0.422 × 5.93)
v² = 5
v = √5
v = 2.24 m/s
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