A block of mass 0.61kg starts from rest at point A and slides down a frictionless hill of height h. At the bottom of the hill it slides across a horizontal piece of track where the coefficient of kinetic friction is 0.40. This section (from points B to C) is 4.73m in length. The block then enters a frictionless loop of radius r= 2.66m. Point D is the highest point in the loop. The loop has a total height of 2r.
a.) What is the minimum speed of the block at point D that still allows the block to complete the loop without leaving the track?
b.) What is the minimum kinetic energy for the block at point C in order to have enough speed at point D that the block will not leave the track?
c.)What is the minimum kinetic energy for the block at point B in order to have enough speed at point D that the block will not leave the track?
d.) What is the minimum height from which the block should start in order to have enough speed at point D that the block will not leave the track?
a) where mg = mv^2/r
b) 1/2 mv^2
c) KE - Ffd = 1/2 mv^2
Where Ff = mu mg
d) mgh = 1/2 mv^2
You can see m cancels in all occasions.
Yes i can see this
mg=mv^2/r
(0.61)9.8=(v^2(0.61))/2.66
v=2.61
To solve these problems, we need to understand the concepts of energy conservation and Newton's second law of motion. We'll use the equations for mechanical energy and centripetal force to find the answers.
a.) To find the minimum speed of the block at point D, we need to consider the forces acting on the block at that point. The only force acting on the block is its weight, which provides the centripetal force necessary for circular motion.
At point D, the weight of the block should equal the centripetal force required to keep the block moving in a circular path. The centripetal force F can be calculated using the equation:
F = mv^2 / r
where m is the mass of the block, v is the speed of the block, and r is the radius of the loop.
Setting the weight equal to the centripetal force:
mg = mv^2 / r
Canceling out the mass "m" and rearranging the equation:
g = v^2 / r
Solving for v:
v = sqrt(gr)
Substituting the values of g (acceleration due to gravity) and r (radius of the loop), we can now calculate the minimum speed of the block at point D.
b.) To find the minimum kinetic energy at point C, we need to consider the forces acting on the block at that point. In addition to its weight, the block also experiences kinetic friction on the horizontal track.
The work done by friction is given by the equation:
W = μk * F * d
where μk is the coefficient of kinetic friction, F is the normal force (equal to the weight of the block), and d is the distance over which the frictional force acts.
The work done by friction will cause a loss in the block's mechanical energy. So we need to subtract the work done by friction from the initial kinetic energy of the block at point B to find the minimum kinetic energy at point C.
c.) Similar to part b), we need to consider the work done by friction on the horizontal track. However, this time, the distance over which the frictional force acts is different (4.73m).
d.) To find the minimum height at which the block should start, we need to consider the conservation of energy. At point D, the block has gravitational potential energy mg(2r) and kinetic energy 1/2 mv^2.
To find the minimum height, we need to equate the total mechanical energy at point D to the initial mechanical energy at point A:
PE(A) + KE(A) = PE(D) + KE(D)
Here, we need to consider the gravitational potential energy at point A, the kinetic energy at point A (which is zero), the gravitational potential energy at point D, and the kinetic energy at point D (which is already known from part a).
By rearranging the equation and substituting appropriate values, we can determine the minimum height at which the block should start.
Now, I'll calculate the answers using the given values.