Given: A 25-kg block is subjected to the force F=100 N. The F= 100 N F= 100 N spring has a stiffness of k = 200 N/m and is unstretched 200 N/m 0.3 m when the block is at A. The contact surface is smooth. Find: Draw the free-body and kinetic diagrams of the block when s-0.4 m. Plan: 1) Define an inertial coordinate system. 2) Draw the block's free-body diagram, showing all external forces. 3) Draw the block's kinetic diagram, showing the inertial force vector in the proper direction.

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eep moving to the right, since the net force to
the right is 68N >0.
The acceleration could be directed to the right if the block is
speeding up or to the left if it is slowing down.
- Is the block accelerating or decelerating?
“Homework”: - What is the equation for Fx(s)?
- What is the maximum distance, smax?

EXAMPLE
Given: A 25-kg block is subjected to
F= 100 N
F= 100 N
the force F=100 N. The
spring has a stiffness of k=
200 N/m and is unstretched
200 N/m
0.3 m
when the block is at A. The
contact surface is smooth.
Find: Draw the free-body and kinetic diagrams of the block
when s-0.4 m.
Plan: 1) Define an inertial coordinate system.
2) Draw the block's free-body diagram, showing all
external forces.
3) Draw the block's kinetic diagram, showing the inertial
force vector in the proper
direction.
ALWAYS LEARNING Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc. PEARSON
All rights reserved.
Transcribed Image Text:EXAMPLE Given: A 25-kg block is subjected to F= 100 N F= 100 N the force F=100 N. The spring has a stiffness of k= 200 N/m and is unstretched 200 N/m 0.3 m when the block is at A. The contact surface is smooth. Find: Draw the free-body and kinetic diagrams of the block when s-0.4 m. Plan: 1) Define an inertial coordinate system. 2) Draw the block's free-body diagram, showing all external forces. 3) Draw the block's kinetic diagram, showing the inertial force vector in the proper direction. ALWAYS LEARNING Dynamics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. PEARSON All rights reserved.
CD Page view
AN Read aloud
The normal force Fn is perpendicular to the contact surface.
There is no friction force since the contact surface is smooth.
When s = 0.4m, the spring's extended length is 0.5m
→4= (0,5 - 0.3)m = 0.2 m → Fs = 200N/m * 0.2m = 40N
W = 245N
How do we determine the resulting
force(s) on the block?
y
F=100 (N)
EFx = F – Fs*4/5 = (100-32)N= 68N
F- 200 A (N)
Fn
40 (N)
£Fy = 0! → Fn (vertical equilibrium)
Fn = W+ Fs*3/5 = (245+24)N = 269N
ALWAYS LEARNING Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc. PEARSON
All rights reserved.
Transcribed Image Text:CD Page view AN Read aloud The normal force Fn is perpendicular to the contact surface. There is no friction force since the contact surface is smooth. When s = 0.4m, the spring's extended length is 0.5m →4= (0,5 - 0.3)m = 0.2 m → Fs = 200N/m * 0.2m = 40N W = 245N How do we determine the resulting force(s) on the block? y F=100 (N) EFx = F – Fs*4/5 = (100-32)N= 68N F- 200 A (N) Fn 40 (N) £Fy = 0! → Fn (vertical equilibrium) Fn = W+ Fs*3/5 = (245+24)N = 269N ALWAYS LEARNING Dynamics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. PEARSON All rights reserved.
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