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)

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