Dynamics Exam2 and Problem Solutions
1. Position time graph of 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=fs
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;
Normal force of the box is;
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,
We draw velocity vs. time graph now. Area under the velocity vs. time graph gives us displacement of the object.
2t-3t: ΔX₃= 2.Vt
We know ΔX₂=5/2.Vt=75m, Vt=30m
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;
Box moves downward with 1m/s2 acceleration.