Answer:
Maximum contact force = 487.4 N
Explanation:
The graph showing force vs time is attached to the solution of this question.
According to the an interpretation of Newton's second law of motion, the magmitude of the change in momentum is equal to the magnitude of the impulse experienced duting this impact.
F.t = Δ(momemtum)
Δ(momemtum) = 0.044 [36 - (-36)] = 0.044 × 72 = 3.168 kgm/s
Meaning, Impulse = 3.168 kgm/s too.
The area under the curve above is the impulse we just obtained.
If the maximum contact force is called F
The impulse = Area under this graph
= (1/2)(F)(4-0) + (F)(7-4) + (1/2)(F)(10-7)
= 2F + 3F + 1.5F = (6.5F) mN.s = (0.0065F) N.s
0.0065F = 3.168
F = (3.168/0.0065) = 487.4 N
Hope this Helps!!!
Answer:8.75 s,
136.89 m
Explanation:
Given
Initial velocity
velocity after 5 s is
Therefore acceleration during these 5 s
therefore time required to stop
v=u+at
here v=final velocity =0 m/s
initial velocity =31.29 m/s
(b)total distance traveled before stoppage
s=136.89 m