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Physical Dynamics

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Can you please do these questions and show all work?

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This solution provides a step-by-step explanation on how to solve the given physics problems.

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** Please see the attached file for the complete solution response **

1.

(please see the attached file)

Acceleration is the rate of change of velocity. At any given instant, it is the slope of the velocity - time graph. As the acceleration - time graph for 0 < t < 1 sec shows constant acceleration of 10 m/s^2, the slope of the v - t graph in this duration must be a straight line with slope 10 m/s^2. For the same reason, slope of v - t graph in the duration 1 < t < 2 sec must be a straight line with slope - 10 m/s^2. During 2 < t < 4, the v - t graph has 0 slope.
(please see the attached file)

Velocity at a given instant is the slope of the displacement - time (s - t) graph at that instant. Further, displacement between two time instants is the area under the v - t graph between the given time instants. Area under v - t graph for different time durations are as follows:

0 < t < 1 : ½ x 1 sec x 10 m/sec = 5 m

1 < t < 2 : ½ x 1 sec x 10 m/sec = 5 m

2 < t < 4 : 0

(please see the attached file)
Forces acting on the two blocks are shown in the fig..

Friction force on block A f_A = μ_AN_A = μ_Am_Agcosθ = 0.2 x 200 x cos60^O = 20 lb

Net force on block A = F_A = m_Agsinθ - μ_Am_Agcosθ - ...

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