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2.71 m/s fast Hans is moving after the collision.
<u>Explanation</u>:
Given that,
Mass of Jeremy is 120 kg ()
Speed of Jeremy is 3 m/s ()
Speed of Jeremy after collision is () -2.5 m/s
Mass of Hans is 140 kg ()
Speed of Hans is -2 m/s ()
Speed of Hans after collision is ()
Linear momentum is defined as “mass time’s speed of the vehicle”. Linear momentum before the collision of Jeremy and Hans is
=
Substitute the given values,
= 120 × 3 + 140 × (-2)
= 360 + (-280)
= 80 kg m/s
Linear momentum after the collision of Jeremy and Hans is
=
= 120 × (-2.5) + 140 ×
= -300 + 140 ×
We know that conservation of liner momentum,
Linear momentum before the collision = Linear momentum after the collision
80 = -300 + 140 ×
80 + 300 = 140 ×
380 = 140 ×
380/140=
= 2.71 m/s
2.71 m/s fast Hans is moving after the collision.
The coefficient of static friction between the chair and the floor is 0.67
Explanation:
Given:
Weight of the chair = 25kg
Force = 165 N (F_applied)
Force = 127 N (F_max)
To find: Coefficient of static friction
The “coefficient of static friction” between a chair and the floor is defined as the ration of maximum force to the normal force acting on the chair
μ_s=
The F_n is equal to the weight multiplied by its gravity
∴=mg
Thus the coefficient of static friction changes as
μ_s=
μ_{s} =
= 0.67