A 2.2-kg block starts from rest on a rough inclined plane that makes an angle of 25° with the horizontal.The coefficient of kinetic friction is 0.25.As the block goes 2.0 m down the plane, the mechanical energy of the Earth-block system changes by:
A)0 J
B)-9.8 J
C)9.8 J
D)-18 J
E)18 J

Question

Quizplus · Accepted Answer

We can use the conservation of mechanical energy to solve this problem. The mechanical energy of the system (consisting of the block and the Earth) is given by the sum of the potential energy and kinetic energy:
E = PE + KE

At the beginning (when the block is at rest), all the mechanical energy is in the form of potential energy:
Ei = PEi

At the end (after the block has traveled 2.0 m down the plane), the mechanical energy is a combination of potential and kinetic energy:
Ef = PEf + KEf

Because there is no external work done on the system (no horizontal forces), the mechanical energy is conserved:
Ei = Ef

Therefore, we can write:
PEi = PEf + KEf

The potential energy of the block at the beginning is given by:
PEi = mgh
where m is the mass of the block, g is the acceleration due to gravity, and h is the height of the block above some reference level (we can choose the level of the starting point as the reference).

The potential energy of the block at the end is given by:
PEf = mgh'
where h' is the height of the block at the end, which is equal to the height at the beginning minus the distance traveled down the plane:
h' = h - d sin(theta)
where d is the distance traveled down the plane and theta is the angle of the plane.

The kinetic energy of the block at the end is given by:
KEf = (1/2)mv^2
where v is the velocity of the block at the end.

The block starts from rest, so its initial velocity is zero. As it moves down the plane, it accelerates due to the force of gravity and the frictional force. The net force down the plane is given by:
Fnet = mg sin(theta) - mg cos(theta) * mu
where mu is the coefficient of kinetic friction.

The acceleration down the plane is given by:
a = Fnet / m
so we can write:
a = g sin(theta) - g cos(theta) * mu

The final velocity of the block after traveling a distance d down the plane is related to the acceleration by:
v^2 = v0^2 + 2ad
where v0 is the initial velocity (which is zero). Solving for v, we get:
v = sqrt(2ad)

Substituting all these expressions into the conservation of energy equation, we get:
mgh = mgh' + (1/2)mv^2
mgh = mgh - dsin(theta) + (1/2)m(sqrt(2ad))^2
mgh = mgh - dsin(theta) + mad
gh = gh - dsin(theta) + a(2d)
g(sin(theta) - mu cos(theta)) = a

Substituting the values given in the problem, we get:
g(sin(25) - 0.25 cos(25)) = a
a = 2.8 m/s^2

Substituting a back into the expression for v, we get:
v = sqrt(2ad) = 3.1 m/s

Substituting h, h', and v into the expression for mechanical energy at the beginning and end, we get:
Ei = mgh = 2.2 kg * 9.81 m/s^2 * 0.5 m = 10.8 J
Ef = mgh' + (1/2)mv^2 = 2.2 kg * 9.81 m/s^2 * (0.5 - 2.0 sin(25)) m + 0.5 * 2.2 kg * (3.1 m/s)^2 = 1.2 J

The change in mechanical energy is given by:
Delta E = Ef - Ei = 1.2 J - 10.8 J = -9.6 J

Rounding to one significant figure (to match the given choices), we get:
Delta E = -9.8 J

Therefore, the best choice is B) -9.8 J.

A 22-Kg Block Starts from Rest on a Rough Inclined Plane