- #1
cukitas2001
- 63
- 0
Hey for anyone up this late, I am having some problems with three questions. all these are to be solved using K_1+U_1=K_2+U_2
1) A mail bag with a mass of 115kg is suspended by a vertical rope of length 7.00m .A)What horizontal force is necessary to hold the bag in a position displaced sideways a distance 3.00m from its initial position?
Take free fall acceleration to be = 9.80m/s^2 . B) How much work is done by the worker in moving the bag to this position?
Take free fall acceleration to be = 9.80m/s^2 .
Part A was easy and i figured it to be 535N but part B is killing me. I tried using W=F*s but it came out wrong and your hinted to use g=9.8m/s^2...any ideas or tips/hints?
2)A spring of negligible mass has force constant = 1800N/m. A) How far must the spring be compressed for an amount 3.20J of potential energy to be stored in it? B) You place the spring vertically with one end on the floor. You then drop a book of mass 1.50kg onto it from a height of 0.900m above the top of the spring. Find the maximum distance the spring will be compressed. Take the free fall acceleration to be = 9.80m/s^2.
Again part A was easy and i found it to be 5.96×10^−2 m but part B is being troublesome. I tried using W=(1/2)K(x_2)^2-(1/2)K(x_1)^2 and made W=mgh which came out to be 13.23J and set this equal to
(1/2)K(x_2)^2-(1/2)K(x_1)^2 and solved for the delta x but got it wrong. Any ides on where my mistakes are?
3)A wooden rod of negligible mass and length 75.0cm is pivoted about a horizontal axis through its center. A white rat with mass 0.510kg clings to one end of the stick, and a mouse with mass 0.150kg clings to the other end. The system is released from rest with the rod horizontal. If the animals can manage to hold on, what are their speeds as the rod swings through a vertical position? Take free fall acceleration to be = 9.80m/s^2
Ok, this one has had me stumped for a few hours now and iv'e tried a couple of different methods. I tried K_1+U_1=K_2+U_2 making K_1 and U_2 equal to zero. eventually i came to the expression of V_2=sqrt(2gr) and used 75cm/2 as r but got the wrong answer. Then i tried substracting the two masses to have an equivalent system where there is only one mass released form rest but agian that took me no where. Any ideas here?
1) A mail bag with a mass of 115kg is suspended by a vertical rope of length 7.00m .A)What horizontal force is necessary to hold the bag in a position displaced sideways a distance 3.00m from its initial position?
Take free fall acceleration to be = 9.80m/s^2 . B) How much work is done by the worker in moving the bag to this position?
Take free fall acceleration to be = 9.80m/s^2 .
Part A was easy and i figured it to be 535N but part B is killing me. I tried using W=F*s but it came out wrong and your hinted to use g=9.8m/s^2...any ideas or tips/hints?
2)A spring of negligible mass has force constant = 1800N/m. A) How far must the spring be compressed for an amount 3.20J of potential energy to be stored in it? B) You place the spring vertically with one end on the floor. You then drop a book of mass 1.50kg onto it from a height of 0.900m above the top of the spring. Find the maximum distance the spring will be compressed. Take the free fall acceleration to be = 9.80m/s^2.
Again part A was easy and i found it to be 5.96×10^−2 m but part B is being troublesome. I tried using W=(1/2)K(x_2)^2-(1/2)K(x_1)^2 and made W=mgh which came out to be 13.23J and set this equal to
(1/2)K(x_2)^2-(1/2)K(x_1)^2 and solved for the delta x but got it wrong. Any ides on where my mistakes are?
3)A wooden rod of negligible mass and length 75.0cm is pivoted about a horizontal axis through its center. A white rat with mass 0.510kg clings to one end of the stick, and a mouse with mass 0.150kg clings to the other end. The system is released from rest with the rod horizontal. If the animals can manage to hold on, what are their speeds as the rod swings through a vertical position? Take free fall acceleration to be = 9.80m/s^2
Ok, this one has had me stumped for a few hours now and iv'e tried a couple of different methods. I tried K_1+U_1=K_2+U_2 making K_1 and U_2 equal to zero. eventually i came to the expression of V_2=sqrt(2gr) and used 75cm/2 as r but got the wrong answer. Then i tried substracting the two masses to have an equivalent system where there is only one mass released form rest but agian that took me no where. Any ideas here?