Question

woman pulls a 6.87 kg suitcase,
initially at rest, with a 29.2 N force
along the handle at 57.7º. Friction
pulls back at 12.7 N, and the
suitcase moves 5.93 m. What is
its final speed?
(Unit = m/s)​

Answer 1

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

Answer 2

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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