HELP +- Principle of Linear Impulse and Momentum Learning Goal: To apply the pri
ID: 1286135 • Letter: H
Question
HELP
+- Principle of Linear Impulse and Momentum Learning Goal: To apply the principle of linear impulse and momentum to a mass to determine the final speed of the mass. A 10-kg, smooth block moves to the right with a velocity of v0 m/s when a force F is applied at time t0 = 0 s. Part A - Constant Forces At time t0 = 0 s, a force is applied to the block. If the block is moving with an initial velocity v0 = 1.25 m/s and the force varies as shown in the graph, where t1 = 1 s, t2 = 2 s, and t3 = 3 s, what is the speed of the block at time t3? Express your answer to three significant figures. v(3 s) = ________ m/s Part B -The Speed of the Block att3 At time t0 = 0 s, a force is applied to the block. If the block is moving with an initial velocity V0 = 3.5 m/s and the force varies as shown in the graph, where t1 = 1.25s, t2 = 2.5 s, and t3 = 3.75 s. What is the speed of the block at time t3? Express your answer to three significant figures. v(3.75 s) = _________________ m/s Part C - The Time it take to Stop the Motion of the Block At time t0 = 0 s, a force is applied to the block. The force varies as shown in the graph. where t1 = 1.25 s, t2 = 2.5 s, and t3 = 3.75 s. If the block is moving with an initial velocity v0 = 5 m/s and the force remains constant at -20 N for all times greater than t3, at what time tf does the block stop moving to the right? Express your answer to three significant figures.Explanation / Answer
Impulse is area under the F-t curve
1) Area= 20*1 +10*1-10*1=20 Kg-m/s
==> Impulse = change in momentum
==> 20 kg-m/s = m(v2-v1) = 10*(v2-1.25)
==> v2= 3.25 m/s <------------ speed at t3
2) Area= 0.5*20*2.5 -0.5*20*1.25 = 12.5 kg-m/s
==> Impulse = change in momentum
==> 12.5 kg-m/s = m(v2-v1) = 10*(v2-3.5)
==> v2= 4.75 m/s <------------ speed at t3
3) Area= 0.5*20*1.25-20*(t-2.5) kg-m/s
to stop final momentum= 0
==> Impulse = change in momentum
==> 0.5*20*1.25-20*(t-2.5) kg-m/s = 0-10*5
==> 62.5-20t = -50
==> 112.5 = 20t
==> t = 5.625 s <------------ time to stop
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