1.) After a tornado, a 0.50 g drinking straw was found embedded 4.5 cm in a tree. Subsequent measurements showed that the tree exerted a stopping force of 70 N on the straw. What was the straw's speed?
2.) Find the work done by a force F= 1.8x + 2.2y N as it acts on an object moving from the origin to the point R= 56x + 31y m.
3.) If the coefficient of kinetic friction is 0.22, how much work do you do when you slide a 50 kg box at constant speed across a 5.8m wide room?
2.) Find the work done by a force F= 1.8x + 2.2y N as it acts on an object moving from the origin to the point R= 56x + 31y m.
3.) If the coefficient of kinetic friction is 0.22, how much work do you do when you slide a 50 kg box at constant speed across a 5.8m wide room?

1)
F = m*a with F = 70 N and m = 0.0005 kg
and
v^2 = v0^2 + 2as with v = 0, and s = 0.045 m
v0^2 =  2as
a = v0^2/(2s) sub into F = m*a
70 = v0^2/(2*0.045) * 0.0005
70*2*0.045/0.0005 = v0^2
v0 = 112.25 m/s < speed of straw

the distance moved by the particle is
√(56^2+31^2) = 64 m
Find the magnitude of the force parallel to the direction of the particle movement vector:
use
F  = (Fvector * particle vector)/(magnitude of particle vector)^2 * particle vector (with * = dot product)
the force vector in direction of the particle movement vector is
F = (1.8, 2.2)⋄(56, 31)/√(56^+31^2)^2*(56, 31)
F = (100.8 + 68.2)/64^2*(56, 31)
F = 0.04126*(56, 31) = (2.310, 1.279)
the magnitude of this force vector is
mF = √(2.31^2 + 1.279^2) = 2.64 N
Work = F*s = 2.64*64 = 169 J

Work = friction force*distance
W = μmg*s
W = 0.22*50*9.81*5.8 J
W = 625.88 J
OG
F = m*a with F = 70 N and m = 0.0005 kg
and
v^2 = v0^2 + 2as with v = 0, and s = 0.045 m
v0^2 =  2as
a = v0^2/(2s) sub into F = m*a
70 = v0^2/(2*0.045) * 0.0005
70*2*0.045/0.0005 = v0^2
v0 = 112.25 m/s < speed of straw

the distance moved by the particle is
√(56^2+31^2) = 64 m
Find the magnitude of the force parallel to the direction of the particle movement vector:
use
F  = (Fvector * particle vector)/(magnitude of particle vector)^2 * particle vector (with * = dot product)
the force vector in direction of the particle movement vector is
F = (1.8, 2.2)⋄(56, 31)/√(56^+31^2)^2*(56, 31)
F = (100.8 + 68.2)/64^2*(56, 31)
F = 0.04126*(56, 31) = (2.310, 1.279)
the magnitude of this force vector is
mF = √(2.31^2 + 1.279^2) = 2.64 N
Work = F*s = 2.64*64 = 169 J

Work = friction force*distance
W = μmg*s
W = 0.22*50*9.81*5.8 J
W = 625.88 J
OG