Question
Photos include a picture of the questions and correct answers to questions 1-5 and freebody diagrams. Please use them to ANSWER QUESTIONS 6-10 ONLY. Thank you very much.
uestion. Show your work. Answers with no work receive no credit. Circle numerica answers and give them to three significant digits. Make sure your answers have correct units and that directions are clearly indicated for vectors. Each question is worth 5 points, for 80 points total. A 300-g block of wood is placed on a level table, 80.0 cm from the edge. A string runs horizontally from the block to a pulley at the edge of the table. The string runs over the pulley and down to a 50.0 g hanger that hangs 50.0 cm above the floor. The system is released from rest and the hanger falls to the floor. The frictional force between the block of wood and the table is 0.250 N 1. Draw a free body diagram of the block of wood just as it starts to move. Be sure to list the objects and interactions Draw a free body diagram of the hanger just as it starts to move. Be sure to list the objects and interactions 2. 3. Determine the tension in the string 4. Determine the acceleration of the hanger 5. What is the speed of the block of wood when the hanger hits the floor? 6. What is the acceleration of the block of wood after the hanger hits the floor? 7. If t-0 is when the block is released, at what time does the block of wood reach the 8. 9. 10. edge of the table? If the mass of the block were doubled, how would this situation be different? Be as detailed as possible If instead the table were frictionless, how would the situation (with the 300 g block) be different? Be as detailed as possible If the table were frictionless, how would the situation be different if you doubled the mass of the block? Compare it to both the small-mass, no-friction case and to the doubled-mass, with-friction case.
Explanation / Answer
Q6
As seen ,we can say once it hits the floor,T=0
so accleration=-fs/m=-0.83 m/s^2
Q7
We know till m=50g hits the floor
a=0.687 m/s^2
time taken to cover 50 cm=1.21 sec
velocity =0.687*1.21=0.828 m/s
now accleration=-0.83 m/s^2
distance to travel=30 cm=0.3 cm
0.3=0.828t-0.5*0.83*t^2
t=0.47 sec
total time=0.47+1.21=1.68 sec
Q8
if mass is doubled
accleration will chnage so everything will change
new a=0.37 m/s^2
so will T change
T=0.05(g-a)=0.4715N
Q9
if it was frictionless
friction force will be zero so new a =g*m/(M+m)=1.4 m/s^2
so will chnage Tension
T=0.42 N
Q10
If both case
f=0
m=600=0.6
a=0.754 m/s^2
T=0.4523