A small cube of mass m1= 2.0 kg slides down a circular and frictionless track of radius R= 0.6 m cut into a large block of mass m2= 5.0 kg as shown in the figure below. The large block rests on a horizontal and frictionless table. The cube and the block are initially at rest, and the cube m1 starts from the top of the path. Find the speed of the cube v1 as it leaves the block. Take g= 10.0 m/s2. Enter your answer in m/s.
Good grief will this 8:01 final never end?
Here you have the countdown Damon,
[timeanddate](dot)com/countdown/generic?p0=1440&iso=20140113T2359&year=2014&month=1&day=13&hour=23&min=59&sec=0&msg=Final%20Exam%20Due&csz=1
Thank you !
Morgen, morgen, nur nicht heute
sagen alle faule leute :)
v=2.9277
hai @mahaboob can u help m for dis
A small cube of mass m1= 2.0 kg slides down a circular and frictionless track of radius R= 0.5 m cut into a large block of mass m2= 5.0 kg as shown in the figure below. The large block rests on a horizontal and frictionless table. The cube and the block are initially at rest, and the cube m1 starts from the top of the path. Find the speed of the cube v1 as it leaves the block. Take g= 10.0 m/s2. Enter your answer in m/s. ??
To find the speed of the cube (v1) as it leaves the block, we can use the conservation of energy principle.
The initial energy of the cube is purely gravitational potential energy, which is given by m1*g*h, where m1 is the mass of the cube, g is the acceleration due to gravity, and h is the vertical height from which the cube starts.
The final energy of the cube is the sum of its kinetic energy and gravitational potential energy.
The kinetic energy of the cube is given by (1/2)*m1*v1^2, where v1 is the velocity of the cube as it leaves the block.
The gravitational potential energy of the cube in this final state is zero, as it is at the lowest point of its motion.
Since there is no friction, the initial gravitational potential energy of the cube is equal to the final kinetic energy of the cube.
Therefore, we have:
m1*g*h = (1/2)*m1*v1^2
Rearranging the equation, we get:
v1^2 = 2*g*h
Substituting the given values:
m1 = 2.0 kg
g = 10.0 m/s^2
h = R (since m1 starts from the top of the path)
v1^2 = 2*10.0*0.6
Calculating the value of v1:
v1^2 = 12.0
Taking the square root of both sides, we get:
v1 = √12.0 ≈ 3.46 m/s
Therefore, the speed of the cube (v1) as it leaves the block is approximately 3.46 m/s.