|Energy Work Power|
|Properties Of Matter|
|Heat Temperature And Thermal Expansion|
|Exams and Problem Solutions|
|--Vectors Exam and Problem Solutions|
|--Kinematics Exams and Problem Solutions|
|--Dynamics Exams and Problem Solutions|
|--Dynamics Exam1 and Problem Solutions|
|--Dynamics Exam2 and Problem Solutions|
|--Work Power Energy Exams and Problem Solutions|
|--Impulse Momentum Exams and Problem Solutions|
|--Rotational Motion Exams and Problem Solutions|
|--Optics Exams and Problem Solutions|
|--Properties of Matter Exams and Problem Solutions|
|--Heat Temperature and Thermal Expansion Exams and Problem Solutions|
|--Electrostatics Exams and Problem Solutions|
|--Electric Current Exams and Problem Solutions|
|--Magnetism Exams and Problem Solutions|
|--Waves Exams 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.
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:
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.