1. A 1N pendulum bob is held at an angle theta from the vertical by a 2N horizontal force. Find the tension in the string supporting the pendulum bob. My answer: 2.236N Ty=mg=1N Tx=2N, T=sqr((2N)^2 + (1N)^2)=sqr(5*N^2)=2.236N Without knowing the angle or length I don't know how else to solve this one. 4. Two blocks weighing 250N and 350N respectively are connect by a string that passes over a massless pulleye. Compute the tension in the string. My answer: 291.4N (this 2T, T=145.7N) I am assuming that the blocks would be moving due to the difference in weight. So, M=35.71kg and m=25.51kg (M+m)a=Mg (35.71+25.51)kg*a=35.71kg(9.8m/s^2) (61.22kg)a=349.96N a=349.96N/61.22kg a=5.72m/s^2 T=m*a=(35.71kg)(9.8-5.72)m/s^2 T=145.7N In the book it stated that the magnitude of the rope tension on a pulleye system like this was 2T so it would be 291.4N, but that 2T is when masses are equal, so would it just be 145.7N? 6. At what angle should the roadway on a curve with a 50m radius be banked to allow cars to negotiate the curve at 12m/s even if the roadway is icy( and friction force is 0)? My answer: @=17.09* V=sqr(Rg(sin@+Us*cos@) if Us = 0 then, V=sqr(Rg(sin@)) 12m/s=sqr(Rg(sin@)) 144(m/s)^2=Rg(sin@) 144(m/s)^2=(50m)(9.8m/s^2)(sin@) 144(m/s)^2=490(m/s)^2(sin@) 144/490=sin@ 0.294=sin@ @=sin^-1(0.294)=17.09* 7. A 5.0kg cart is moving horizontally at 6.0m/s. In order to change its speed to 10.0m/s, what must the net work done on the cart be? My answer: 160J change in K=W=Kf-Ki 1/2mVf^2 - 1/2mVo^2 1/2(5.0kg)(10.0m/s)^2 - 1/2(5.0kg)(6.0m/s)^2 2.5kg(100.0m^2/s^2) - 2.5kg(36.0m^2/s^2) 250J-90J= 160J 8. A person holds an 80N weight, 2 meters above the floor for 30 seconds, what is the power required in watts to do this? My answer: 4800watts Wg= -mgd Wg= -80N*2m Wg=-160J Wf=-Wg=160J 1watt=1J/s 30*160J/s=4800J/s=4800watts 10. The speed of a 4.0N hockey puck, sliding on the ice surface decreases at the rate of 0.61m/s^2. What is the coefficient of friction between the puck and the ice? My answer: Uk=0.062 a=-0.61m/s^2 a=-Uk*g -0.61m/s^2= -Uk(9.8m/s^2) -Uk= (-0.61m/s^2)/(9.8m/s^2)=-0.062 Uk=0.062 I think that I did these problems right but you never know, that's why I'd appreciate any feed back. #2,3, and 5 I am still working on and are posted in another thread, and #9 was ommited b/c it's a duplicate. Thanks much.