Please answer the following question completely & correctly. (Please wrtie neatl
ID: 1582887 • Letter: P
Question
Please answer the following question completely & correctly. (Please wrtie neatly & show all work)
1. The following includes materail we learned in class (energy, newtons 2nd law, & radius):
A. In the system below, the rope and pulley have negligible mass, and the pulley is frictionless. The two blocks have masses of 6.0 kg and 4.0 kg, as shown, and the coefficient of kinetic friction between the 6.0 kg block and the tabletop is k 0.230. The blocks are released from rest. 6.0 kg ,-0.230 4.0 kg - Use energy methods to calculate the speed of the 4.0 kg block after it has descended 1.2 m. Use Newton's second law to calculate the speed of the 4.0 kg block after it has descended 1.2 m B. A block of mass m is released on a frictionless track at a height h Find an expression for the minimum value of h that wil allow the block to circle a loop-the-loop with a radius of R, as shown above, without falling off. If the block goes up a hill with a height of 0.75h after the loop-the-loop, what will its speed be at the top of that hill?Explanation / Answer
(A) energy method,
Work done by gravity + work done by friction = change in KE
(4 x 9.8 x 1.2) + (-0.230 x 6 x 9.8 x 1.2) = (6 + 4)v^2 /2 - 0
v = 2.48 m/s .....Ans
Using NEwton's 2nd law,
4g - T = 4 a
T - 0.230 x 6g =6 a
adding, 4g - 1.38g = 10 a
a = 2.57 m/s^2
vf^2 - vi^2 = 2 a d
v^2 - 0^2 = 2(2.57)(1.2)
v = 2.48 m/s ......Ans
(B) at top point in loop,
N + m g = m v^2 / r
{making N = 0}
v = sqrt(r g )
Now applyng energy conservation,
Pei + KEi = PEf + KEF
m g h + 0 = m g (2 r) + m (sqrt(rg))^2 / 2
g h = 2 r g + ( r g / 2)
h = 2.5R .....Ans
m g (2.5R) + 0 = m g (0.75 x 2.5 R) + m v^2 /2
2.5 g R = 1.875 g R + v^2 / 2
v = sqrt[1.25 g R ]