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## Dynamics Exam2 and Problem Solutions

**Dynamics Exam2 and Problem Solutions
**

**1. **Position time graphof the box is given below. Find the friction constant between box and surface? (g=10m/s2)

Slope of the graph gives us velocity of the box. Since the slope of the position time graph is constant, velocity of the box is also constant. As a result, acceleration of the box becomes zero.

**Fnet=F-fs=m.a=0**

**Fnet=fs**

**fs=12**

**k.mg=12**

**k.3.10=12**

**k=0,4**

**2.** If the acceleration of the system given below is 3m/s2, find the friction constant between box and surface. (sin370=0,6, cos370=0,8, sin450=cos450=√2/2)

Free body diagrams of the system are given below.

Acceleration of the 10 kg box is 2m/s2. Thus, net force acting on this box is;

**Fnet=m.a**

**Fnet=10.2=20N**

**Normal force of the box is;**

**N=100+40-60=80N**

**Fnet=80-40-Ffriction**

**20=80-40-k.80**

**k.80=20 **

**k=1/4**

**3. **Net force vs. time graph of object is given below. If displacement of this object between t-2t is 75m, find the displacement of the object between 0-3t.

We draw acceleration vs. time graph using force vs time graph of the object.

Area under the graph gives velocity.

**If we say at=V then,**

**Vt=2V**

**V2t=3V**

**V3t=V**

We draw velocity vs. time graph now.

Area under the velocity vs. time graph gives us displacement of the object.

**0-t: ΔX1=2Vt/2=Vt**

**t-2t: ΔX2=5/2.Vt**

**2t-3t: ΔX3= 2.Vt**

**We know ΔX2=5/2.Vt=75m, Vt=30m**

**Total displacement=ΔX1+ΔX2+ΔX3=Vt+5/2.Vt+2Vt**

**Total displacement=30+75+2.30=165m**

**4.** An object is pulled by constant force F from point A to C. Draw the acceleration vs. time graph of this motion. (F>mg.sinθ and surface is frictionless.)

Motion of the box between points A to B:

**F.cosθ=m.a1**

When the object gets closer to point B, θ becomes larger, and value of cosθ decreases. Thus, a1 decreases between the points A -B.

Motion between points B-C

Net force between points B and C is constant. Thus, a2 is also constant. Acceleration vs. time graph of the box is given below;

**5.** System in the given picture below, box moves under the effect of applied force and gravity with 1m/s2 acceleration. Find the friction constant between the box and surface.

Free body diagram of the system is given below;

Forces acting on the box perpendicularly;

**30+80=110N**

Box moves downward with 1m/s2 acceleration.

**Fnet=m.a**

**60-40-Ffriction=10.1**

**20-k.110=10**

**10=110k**

**k=1/11**

Author: Mrs. Şerife (Erden) SARICA

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